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 147043105X, 9781470431051

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Table of contents :
Cover
Title page
Contents
Foreword
Acknowledgement
Preface
Chapter 0. Introduction
Chapter 1. A basis for \calL
Chapter 2. A basis for \calU
Chapter 3. The Chevalley Groups
Chapter 4. Simplicity of ?
Chapter 5. Chevalley Groups and Algebraic Groups
Chapter 6. Generators and Relations
Chapter 7. Central Extensions
Chapter 8. Variants of the Bruhat Lemma
Chapter 9. The Orders of the Finite Chevalley Groups
Chapter 10. Isomorphisms and Automorphisms
Chapter 11. Some Twisted Groups
Chapter 12. Representations
Chapter 13. Representations Continued
Chapter 14. Representations Completed
Appendix on Finite Reflection Groups
I. Preliminaries
II. The Function ?
III. A Fundamental Domain for ?
IV. Generators and relations for ?
V. Appendix
Bibliography
Index
Back Cover

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UNIVERSITY LECTURE SERIES VOLUME 66

Lectures on Chevalley Groups Robert Steinberg

American Mathematical Society

Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

https://doi.org/10.1090//ulect/066

Lectures on Chevalley Groups

Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

UNIVERSITY LECTURE SERIES VOLUME 66

Lectures on Chevalley Groups Robert Steinberg Notes prepared by John Faulkner and Robert Wilson

American Mathematical Society Providence, Rhode Island

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EDITORIAL COMMITTEE Jordan S. Ellenberg William P. Minicozzi II (Chair)

Robert Guralnick Tatiana Toro

2010 Mathematics Subject Classification. Primary 20G15; Secondary 14Lxx.

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Library of Congress Cataloging-in-Publication Data Names: Steinberg, Robert, 1922- | Steinberg, Robert, 1922- Works. 1997. Title: Lectures on Chevalley groups / Robert Steinberg ; notes prepared by John Faulkner and Robert Wilson. Description: Providence, Rhode Island : American Mathematical Society, [2016] | Series: University lecture series ; volume 66 | Part of: Robert Steinberg, collected papers (Providence, R.I. : American Mathematical Society, 1997). | Includes bibliographical references and index. Identifiers: LCCN 2016042277 | ISBN 9781470431051 (alk. paper) Subjects: LCSH: Chevalley groups. | AMS: Group theory and generalizations – Linear algebraic groups and related topics – Linear algebraic groups over arbitrary fields. msc | Algebraic geometry – Algebraic groups – Algebraic groups. msc Classification: LCC QA179 .S74 2016 | DDC 512/.5–dc23 LC record available at https://lccn. loc.gov/2016042277

Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Permissions to reuse portions of AMS publication content are handled by Copyright Clearance Center’s RightsLink service. For more information, please visit: http://www.ams.org/rightslink. Send requests for translation rights and licensed reprints to [email protected]. Excluded from these provisions is material for which the author holds copyright. In such cases, requests for permission to reuse or reprint material should be addressed directly to the author(s). Copyright ownership is indicated on the copyright page, or on the lower right-hand corner of the first page of each article within proceedings volumes. c 2016 by the American Mathematical Society. All rights reserved.  The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America. ∞ The paper used in this book is acid-free and falls within the guidelines 

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Contents Foreword

vii

Acknowledgement

ix

Preface

xi

Chapter 0. Introduction

1

Chapter 1. A basis for L

5

Chapter 2. A basis for U

11

Chapter 3. The Chevalley Groups

19

Chapter 4. Simplicity of G

33

Chapter 5. Chevalley Groups and Algebraic Groups

37

Chapter 6. Generators and Relations

43

Chapter 7. Central Extensions

47

Chapter 8. Variants of the Bruhat Lemma

61

Chapter 9. The Orders of the Finite Chevalley Groups

77

Chapter 10. Isomorphisms and Automorphisms

85

Chapter 11. Some Twisted Groups

99

Chapter 12. Representations

117

Chapter 13. Representations Continued

129

Chapter 14. Representations Completed

137

Appendix on Finite Reflection Groups I. Preliminaries II. The Function N III. A Fundamental Domain for W IV. Generators and relations for W V. Appendix

149 149 151 152 153 154

Bibliography

155

Index

159

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Foreword In 1997 the American Mathematical Society published Robert Steinberg, The Collected Papers, which contains forty key publications selected by the author. One notable omission, however, was his Lectures on Chevalley Groups, which has only appeared in mimeograph form, printed by the Yale Mathematics Department. It was my uncle’s constant desire to publish a revised version of this work. As the author’s namesake, it is a privilege to assist in the completion of this posthumous edition: Robert Steinberg was a devoted uncle and source of inspiration to all of his nieces and nephews. The current volume incorporates a new introductory chapter, corrections contributed by astute readers of the original edition, and changes that Robert Steinberg penciled into his personal copies. Footnotes marked by “()” are English translations of footnotes from the 1975 Russian edition of the work (edited by A.A. Kirillov). This edition would not be possible without the generous assistance of Christopher Drupieski, who painstakingly transcribed the original mimeographed edition into LATEX, nor without the editorial contributions of my uncle’s dear colleague and friend Raja V. S. Varadarajan. I would also like to acknowledge the editorial staff of the American Mathematical Society, and especially Sergei Gelfand for creating the bibliography and the index, and for translating the editorial footnotes from the Russian edition. Since its original publication, several books have appeared that may help the reader approach these lectures, including Roger W. Carter, Simple Groups of Lie Type, Wiley, New York, 1989; James E. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer, New York, 1972; and V. S. Varadarajan, Lie Groups, Lie Algebras, and Their Representations, Springer, New York, 1984. Robert R. Snapp

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Acknowledgement The notes are dedicated to my wife, Maria. They might also have been dedicated to the Yale Mathematics Department typing staff whose uniformly high standards will become apparent to anyone reading the notes. Needless to say, I am greatly indebted to John Faulkner and Robert Wilson for writing up a major part of the notes. Finally, it is a pleasure to acknowledge the great stimulus derived from my class and from many colleagues, too many to mention by name, during my stay at Yale.

Robert Steinberg June 1968

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Preface These notes presuppose the theory of complex semisimple Lie algebras through the classification, as may be found in the books of E. B. Dynkin, N. Jacobson, or J.-P. Serre, or in the notes of S´eminaire Sophus Lie. An appendix dealing with the most frequently needed results about finite reflection groups and root systems has been included. The reader is advised to read this part first, which can be done rather quickly.

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https://doi.org/10.1090//ulect/066/01

CHAPTER 0

Introduction The groups now called Chevalley groups were introduced in a ground-breaking paper [14] by Claude Chevalley, perhaps the leading figure in the theory of Lie groups and algebraic groups in the middle of the 20th century. In their structure, Chevalley groups look like Lie groups, except that the usual number fields of analysis, the reals and the complexes, can be replaced by arbitrary fields, including finite ones. In fact, one of Chevalley’s obvious goals in writing this paper was to obtain new families of finite simple groups corresponding to the exceptional Lie groups of type E7 and E8 , and almost immediately, it led to a flurry of activity, some of it by the present author, in which variants of his method led to yet other new families of finite simple groups, which, as it turned out, rounded out the final classification of these groups, except for a handful of “sporadic” ones which were found, one at a time, in the next few years. In these notes we study these groups, with their definitions somewhat extended in a number of different areas, listed in the Table of Contents. Since our starting point is the theory of semisimple and simple Lie algebras over C, a knowledge of this subject through the classification would be very helpful to the reader. Fortunately, there are several very good books on the subject, among them Jabcobson [35] and Humphreys [31]. Since even those people who have been through the classification might not have all of the relevant facts at their fingertips, I recall the main ones now, in this preliminary section, to provide an “in house” basis for our later discussion. A Lie algebra is, for us, a finite-dimensional vector space L over some field, to start with C, equipped with a bilinear multiplication {X, Y } → [X, Y ], such that (1) (2)

[X, X] = 0,

and

[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0, for all X, Y, Z in L.

From (1) it follows that [X, Y ] = −[Y, X], and vice versa, since 2 = 0 in C. The relation (2) is called the Jacobi identity. Important examples of Lie algebras are obtained from associative algebras, such as the algebra of linear transformations on a vector space, by introducing the “bracket multiplication” X, Y → XY − Y X in terms of the original multiplication in the algebra. An ideal in L is a subspace I such that [X, Y ] ∈ I for all X ∈ L and Y ∈ I. L is called abelian if [X, Y ] = 0 for all X, Y ∈ L, and semisimple if it has no nonzero abelian ideal. It is then, essentially uniquely, the direct sum of simple ideals (each of which is nonzero and has no ideals other than itself and 0). The simple Lie algebras over C were first classified in the 1890’s (but not called “Lie algebras” at the time), first by Wilhelm Killing, and then, in his thesis by Elie Cartan who corrected some serious mistakes that Killing had made in his proofs and otherwise improved the development. We Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 1

2

0. INTRODUCTION

now sketch the classification, as it is done today, for example in Jacobson’s [35] and Humphreys’ [31] books, following mainly Cartan. We first note that Jacobi’s identity has (at least) two immediate consequences. First, if ad : L → L denotes left multiplication, (ad X)Y = [X, Y ] for all X, Y ∈ L, then ad[X, Y ] = ad X ad Y − ad Y ad X (as we see by applying both sides to an arbitrary Z), so that L can be viewed as an algebra of linear transformations if that is convenient. Second, (ad X)[Y, Z] = [(ad X)Y, Z] + [Y, (ad X)Z], so that each ad X is a derivation, and hence has an exponential which is an automorphism. Now let H be a Cartan subalgebra of L, one definition of which is a maximal subalgebra whose adjoint action on L is diagonalizable; i.e., it is represented by diagonal matrices relative to a suitable linear basis of L, and hence is abelian. Then L breaks down under ad H to a direct sum of H and a number of 1-dimensional subspaces Lα , corresponding to distinct linear functions α on H called roots (coming from characteristic root, as in linear algebra) such that (3)

[H, X] = α(H),

for all H ∈ H and X ∈ Lα .

One also has (4)

[Lα , Lβ ] = Lα+β ,

if α and β are roots such that α + β is also. The inclusion of the left side in the right follows from Jacobi’s identity applied to H ∈ H, X ∈ Lα and Y ∈ Lβ . To go farther, in fact to get even this far, one has to introduce the Killing form. This is the bilinear form κ on L defined by (5)

κ(X, Y ) = tr(ad X ad Y ),

where, tr is the trace, for all X, Y ∈ L. It is symmetric and nondegenerate and satisfies, as is easily checked, (6)

κ(X, [Y, Z]) = κ([X, Y ], Z),

for all X, Y , and Z. Written as κ((ad Y )X, Z) + κ(X, (ad Y )Z) = 0, this indicates the invariance of κ(X, Z) under all of the derivations ad Y . (It is “constant” because all of its derivatives are 0.) It is also nondegenerate: by (6) all of the X ∈ L such that κ(X, L) = 0 are seen to form an ideal, which must be 0 or L since L is simple. It can not be all of L since then L would be solvable by a theorem of Cartan and hence not be simple. Thus the ideal is 0 and κ is nondegenerate. It follows from this that if α is any root, then −α is also. Otherwise for any root β and any H ∈ H, ad Xα · ad Xβ and ad Xα · ad H would map every root space (including H = L0 ) into a different one and thus have trace 0. (Viewed as a matrix, all of its diagonal elements would be 0.) Thus L−α exists and ad Lα · ad L−α = 0. This is a crucial point in the development since then Xα ∈ Lα , X−α ∈ L−α , and Hα ∈ H can be found that span a simple Lie algebra isomorphic to sl2 (2 × 2 matrices of trace 0) with its usual basis, and it is the action of these subalgebras (one for each pair of roots ±α) on the whole algebra that unravel its innermost secrets. The argument can also be used to show that κ remains nondegenerate when restricted to H. It can therefore be transferred to H∗ , the linear dual of H, thus: For a fixed H ∈ H, κ(H, H1 ) is a linear function of H1 ∈ H, say H ∗ , so that H ∗ (H1 ) = κ(H, H1 ) for all H1 ∈ H. The map H → H ∗ is a linear isomorphism of H onto H∗ which can be used to transfer κ to a nondegenerate symmetric bilinear form (·, ·) on H∗ such that (H1∗ , H2∗ ) = κ(H1 , H2 ) for all H1 , H2 ∈ H. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

0. INTRODUCTION

3

The roots, which lie in H∗ , now come to the fore. It turns out that there are l roots (l = dim H = rank L) such that every root is an integral linear combination of these “simple” roots in which all of the nonzero coefficients have the same sign (all positive or all negative). Hence every linear relation of the roots over C is a consequence of the relations over Z. Further the form (·, ·) restricted to the space spanned by the roots over Q or R is positive definite, and the set of roots, which we call Σ, is invariant under the reflection in the hyperplane perpendicular to any of them: for any roots α and β, wα β = β − 2(β, α)/(α, α) · α is also a root. Related to this is the fact that the numbers 2(β, α)/(α, α), which we shall write as β, α, are all integers, called Cartan integers; and the (finite) group generated by the reflections is written W and called the Weyl group. Finally, the classification of the simple Lie algebras over C comes down to the classification of these root systems: two such algebras are isomorphic if and only if their root systems are isometric; any collection of elements in a Euclidean space which satisfies the properties just enumerated for the roots and does not split into two mutually orthogonal sets is, up to a change in scale, the root system of a simple Lie algebra over C. We close with a brief discussion of the actual simple Lie algebras. First, there is L = sln (n ≥ 2), the Lie algebra of n × n matrices of trace 0 with multiplication [X, Y ] = XY −Y X. Here one choice of H is the subalgebra of its diagonal elements, so that L has rank n − 1, and is usually labelled An−1 . The roots are the maps diag(a1 , a2 , . . . , an ) → ai − aj (1 ≤ i, j ≤ n, i = j), for which the root spaces are {CEij }, Eij the matrix unit which has a 1 in the (i, j) position and zeros in all other positions. An−1 may be viewed as the Lie algebra of the Lie group G = SLn (C) of n × n matrices of determinant 1, since to get the tangents at the unit element 1, one just has to find the matrices 1 +X, with  an infinitesimal such that 2 = 0, that satisfy det(1 + X) = 1, i.e., 1 + tr X ·  = 1, i.e., tr X = 0. Other examples are obtained from the orthogonal groups which stabilize a nondegenerate symmetric bilinear form or one in which the form is skew symmetric. In the first, the underlying space has arbitrary dimension; in the second, it must be even. If A is the matrix of the form, then an element of the group, x, must satisfy, xAxt = A, i.e, x must be invariant under the map x → A(x−1 )t A−1 , which, for x = 1 + X becomes: X must be invariant under X → −AX t A−1 . To get an example where the roots and the root spaces can be worked out explicity, try taking A to be a square matrix supported by the off-diagonal (which goes from the upper right hand corner of a matrix to its lower left hand corner, with the entries on this diagonal all 1 for the orthogonal groups (and their Lie algebras), and alternately 1 and −1 for the symplectic groups, for which n must be even. The resulting Lie algebras will all have a Cartan subalgebra on the main diagonal and the “positive” root vectors above the diagonal. Aside from these “classical” Lie algebras there are just 5 other simple algebras over C called E8 , E7 , E6 , F4 and G2 , with the subscript indicating the rank of the algebra (the dimension of a Cartan subalgebra). The group corresponding to G2 is the group of automorphisms of an exotic 8-dimensional algebra called the Cayley numbers, and that corresponding to E8 appears frequently in the Science Section of The New York Times in the ongoing efforts of physicists (and mathematicians) to explain the mysteries of the universe.

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https://doi.org/10.1090//ulect/066/02

CHAPTER 1

A basis for L We start with some basic properties of semisimple Lie algebras over C, and establish some notation to be used throughout. The assertions not proved here are proved in the standard books on Lie Algebras, e.g., those of Jacobson [35], Humphreys [31], or the S´eminaire “Sophus Lie” [52]. Let L be a semisimple Lie algebra over C, and H a Cartan subalgebra of L. Then H is necessarily abelian and L = H ⊕ α=0 Lα where α ∈ H∗ and Lα = {X ∈ L : [H, X] = α(H)X, ∀ H ∈ H} . Each Lα (α = 0) has dimension 1. Note that H = L0 . The α’s are linear functions on H, called roots. We adopt the convention that Lγ = 0 if γ is not a root. Then [Lα , Lβ ] ⊆ Lα+β . The rank of L equals dimC H = l, say. The roots generate H∗ as a vector space over C. ∗ , the vector space over Q generated by the roots. Then dimQ V = Write V for HQ l. Let γ ∈ V . Since the Killing form is nondegenerate on H there exists an Hγ ∈ H such that (H, Hγ ) = γ(H) for all H ∈ H. Define (γ, δ) = (Hγ , Hδ ) for all γ, δ ∈ V . This is a symmetric, nondegenerate, positive bilinear form on V . Denote the collection of all roots by Σ. Then Σ is a subset of the nonzero elements of V satisfying: (0) Σ generates V as a vector space over Q. (1) α ∈ Σ ⇒ −α ∈ Σ and kα ∈ / Σ for every k an integer = ±1. (2) 2(α, β)/(β, β) ∈ Z for all α, β ∈ Σ. (Write α, β = 2(α, β)/(β, β). These are called Cartan integers.) (3) Σ is invariant under all reflections wα (α ∈ Σ) (where wα is the reflection in the hyperplane orthogonal to α, i.e., wα v = v − 2(v, α)/(α, α) α). Thus Σ is a root system in the sense of Appendix I. Conversely, if Σ is any root system satisfying condition (2), then Σ is the root system of some Lie algebra. The group W generated by all wα is a finite group (Appendix I.6) called the Weyl group. If {α1 , . . . , αl } is a simple system of roots (Appendix I.8), then W is generated by the wαi (i = 1, . . . , l) (Appendix I.16) and every root is congruent under W to a simple root (Appendix I.15). Lemma 1. For each root α, let Hα ∈ H be such that (H, Hα ) = α(H) for all H ∈ H. Define Hα = 2/(α, α)Hα and Hi = Hαi (i = 1, . . . , l). Then each Hα is an integral linear combination of the Hi (the H’s for the simple roots). Hj

Proof. Write wi for wαi ∈ W . Define an action of W on H by wi Hj = − αj , αi  Hi . (This is just the adjoint action on H = (H∗ )∗ , coming from the

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1. A BASIS FOR L

6

action of W on H∗ = V .) Then w i Hj

= = = = =

2 wi Hj (αj , αj ) 2(αi , αj )  2 2 Hj − H (αj , αj ) (αj , αj ) (αi , αi ) i 2(αi , αj )  2 Hj − H (αi , αi ) (αj , αj ) i Hj − αi , αj  Hi (an integral linear combination) Hwi αj .

Thus since the wi generate W , wHj is an integral linear combination of the Hi for all w ∈ W . Now if α is an arbitrary root then α = wαj for some w ∈ W and some j. Then Hα = Hwαj = wHαj = wHj = an integral linear combination of  the Hi . For every root α choose Xα ∈ Lα , Xα = 0. If α + β = 0, define Nα,β by [Xα , Xβ ] = Nα,β Xα+β . Set Nα,β = 0 if α + β is not a root. If α and β are roots the α-string of roots through β is the sequence β − rα, . . . , β, . . . , β + qα where β + iα is a root for −r ≤ i ≤ q but β − (r + 1)α and β + (q + 1)α are not roots. Lemma 2. The Xα can be chosen so that: (a) [Xα , X−α ] = Hα for all α, (b) If α and β are roots, β = ±α, such that α + β is also a root, and β − rα, . . . , β, . . . , β + rα is the α-string of roots through β, then 2

2

2 Nα,β = q(r + 1) |α + β| / |β| .

Proof. See the first part of the proof of Theorem 10 in Jacobson [35, p. 147].  2

2

Lemma 3. 1 If α, β and α + β are roots, then q(r + 1) |a + b| / |β| = (r + 1)2 . Proof. We use two facts: (∗) r − q = β, α . (For wα maps β − rα to β + qα so β + qα = wα (β − rα) = β − rα − 2(β − rα, α)/(α, α)α = β − β, α α + rα.) (∗∗) In the α-string of roots through β at most two root lengths occur. (For if V  is the vector space over Q generated by α and β and Σ = Σ ∩ V , then Σ is a root system and every root in the α-string of roots through β belongs to Σ . Now V  is 2 dimensional; so a system of simple roots for Σ has at most two elements. Since every root in Σ is conjugate under the Weyl group of Σ to a simple root, Σ and hence the α-string of roots through β has at most two root lengths.) 1 An alternate proof of Lemmas 2 and 3 was discovered in the author’s notes. It is presented as an Appendix to this chapter on page 8. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

1. A BASIS FOR L

7

We must show that q |α + β|2 / |β|2 = r + 1. Now by (∗): r+1−q

|α + β|2 |β|

2

=

q + β, α + 1 − q

=

β, α + 1 − q Ç

=

|α|2 |β|

2

(α + β, α + β) (β, β) − q α, β

(β, α + 1) 1 − q

|α|

2

|β|2

å .

Set A = β, α + 1 and B = 1 − q |α|2 / |β|2 . We must show A = 0 or B = 0. If |α| ≥ |β| then |β, α| = 2 |(β, α)| / |α|2 ≤ 2 |(β, α)| / |β|2 = |α, β|. By Schwarz’s inequality β, α α, β = 4(α, β)2 / |α|2 |β|2 ≤ 4 with equality if and only if α = kβ. Since α and β are roots with α = ±β we have α = kβ so β, α α, β < 4. Then since |β, α| ≤ |α, β| we have β, α = −1, 0 or 1. If β, α = −1, then A = 0. If β, α ≥ 0 then |β + 2α| > |β + α| > |β|. Since there are only two root lengths β + 2α is not a root and hence q = 1. Since |β + α| > |α| and |β + α| > |β| and at most two root lengths occur, |α| = |β|. Hence B = 0. If |α| < |β|, then |α + β| ≤ |β| (since otherwise three root lengths would occur). Hence (α, β) < 0 so α, β < 0. Then |β − α| > |β| > |α| so β − α is not a root and hence r = 0. As above α, β β, α < 4 and |α, β| < |β, α| so α, β = −1, 0 or 1. Hence α, β = −1. Then by (∗) q = − β, α = β, α / α, β = |β|2 / |α|2 . Hence B = 0.  We collect these results in: Theorem 1. The Hi (i = 1, . . . , l) chosen in Lemma 1 together with the Xα chosen in Lemma 2 form a basis for L relative to which the equations of structure are as follows (and, in particular, are integral): (a) [Hi , Hj ] = 0. (b) [Hi , Xα ] = α, αi  Xα (c) [Xα , X−α ] = Hα = an integral linear combination of the Hi . (d) [Xα , Xβ ] = ±(r + 1)Xα+β if α + β is a root. (e) [Xα , Xβ ] = 0 if α + β = 0 and α + β is not a root. Proof. (a) holds since H is abelian. (b) holds since [Hβ , Xα ] = α(Hβ )Xα = α, β Xα . (c) follows from the choice of the Xα and the Hi from Lemma 1. (d) follows from Lemma 2(b) and Lemma 3. (e) holds since [Lα , Lβ ] = 0 if α + β is not a root.  Remarks. (a) Such a basis is called a Chevalley basis. It is unique up to sign changes and automorphisms of L. (b) Xα , X−α and Hα span a 3-dimensional subalgebra isomorphic to sl2 (2 × 2 matrices of trace 0), with [X, Y ] = XY − Y X, in terms of ordinary multiplication, the standard way of converting an associative algebra into a Lie algebra, ï ò ï ò ï ò 0 1 0 0 1 0 X−α ↔ Hα ↔ . Xα ↔ 0 0 1 0 0 −1 (c) As an example let L = sll+1 , the set of (l + 1) × (l + 1) matrices of trace 0. Then H = {diagonal matrices} is a Cartan subalgera. For i, j = 1, . . . , l + 1, i = j, Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

8

1. A BASIS FOR L

define α = α(i, j) by α(i, j) : diag(a1 , . . . , al+1 ) → ai − aj . Then the α(i, j) are the roots. Let Ei,j be the matrix with 1 in the (i, j) position and 0 elsewhere. Then Xα = Ei,j , X−α = Ej,i , Hα = Ei,i − Ej,j and Hi = Ei,i − Ei+1,i+1 . Exercise. If only one root length occurs then all coefficients in (d) of Theorem 1 are ±1. Otherwise ±2 and ±3 can occur. Appendix Lemma (Alternate form of Lemmas 2 and 3). The root elements Xα ∈ Lα can be chosen so that [Xα , X−α ] = Hα for all α ∈ Σ. (a) This can be done so that whenever α and β are roots such that α + β is also a root and we we set [Xα , Xβ ] = Nα,β Xα+β , then N−α,−β = −Nα,β . (b) It is then true that Nα,β = ±(r + 1), r being the nonnegative integer such that β − rα is a root, but β − (r + 1)α is not. Proof. (a) Multiplication by −1 on H∗ acts as an isometry on the roots therein, and hence, by the uniqueness theorem of the classification, is induced by an automorphism σ of L which also acts as −1 on H and interchanges the root spaces Lα and L−α for every root α: σXα = cX−α and σX−α = dXα for some c and d in C∗ . If we apply σ to the relations [Xα , X−α ] = Hα = −H−α = −[X−α , Xα ], we see that cd = 1 (so that σ is actually an involution). If we choose a ∈ C∗ so that a2 c = −1 (and hence a−2 d = −1), and then replace Xα by aXα and X−α by a−1 X−α , we achieve the normalization c = d = −1. Then if α and β are roots such that α + β is also, σ applied to the relation [Xα , Xβ ] = Nα,β Xα+β yields [Xα , X−β ] = Nα,β (−X−α−β ), so that N−α,−β = −Nα,β . (Observe that in this case L = sln , σ is just the map that sends every matrix to its negative transpose.) (b) This will be done in several steps. With α and β as given, let {β + nα : −r ≤ n ≤ q} be the α-chain of roots through β. (1) q − r = β, α. This expresses the fact that the chain is invariant under the reflection wα . (2) [X−α , [Xα , Xβ ]] = q(r + 1)Xβ , that is, N−α,α+β Nα,β = q(r + 1). We prove this by induction on r, using Jacobi’s identity that [X−α , [Xα , Xβ ]] = [[X−α , Xα ], Xβ ] + [Xα , [X−α , Xβ ]], with the first summand equal to −[Hα , Xβ ] = − β, α Xβ . If r = 0, i.e., β − α is not a root, the second term is 0 and the first term equals qXβ by (1), yielding (2) in this case. Now assume that r > 0   ] for some Xβ−α in Lβ−α . Then so that β − α is a root. Write Xβ = [Xα , Xβ−α  the first term yields (q − r)Xβ again and the second term [Xα , [X−α , [Xα , Xβ−α ]]]  which equals [Xα , (q + 1)rXβ−α ], using the induction assumption with β replaced by β − α and hence q by q + 1 and r by r − 1. The sum of the two terms is thus (q − r + (q + 1)r)Xβ = q(r + 1)Xβ , as required. (3) We now complete the proof of (b). Write r = r(α, β) to indicate its dependence on α and β. Then q = r(−α, α + β) + 1 since (α + β) − (q − 1)(−α) is a root but (α + β) − q(−α) is not. If we set γ = −α − β so that α + β + γ = 0, the equation in (2) becomes N−α,−γ Nα,β = (r(−α, −γ) + 1)(r(α, β) + 1). We now use the fact that r(ρ, σ) = r(σ, ρ) for any two roots, which we prove below in (4). Using this fact, the result of (a), and the fact that Nρ,σ = −Nσ,ρ and Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

1. A BASIS FOR L

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r(ρ, σ) = r(−ρ, −σ) always, our equation can be written Nγ,α Nα,β = (r(γ, α) + 1)(r(α, β) + 1). If we permute α, β, and γ cyclically, we get two more equations like this one. If we then multiply the first two equations together and divide by the third, we get 2 = (r(α, β) + 1)2 . Thus, Nα,β = ±(r(α, β) + 1), as required. Nα,β (4) It remains to prove that r(α, β) = r(β, α) for any roots α and β such that α + β is also a root. If β − α is not a root, then α − β is not a root, so that r(α, β) and r(β, α) are both 0 in this case. Assume then that β − α is a root. Since β + α is a root also and neither the sum nor the difference of these two roots is a root, they must be orthogonal. Thus wβ−α sends β + α, β − α to β + α, −(β − α), respectively, and thus (take sums and difference) interchanges α and β, and hence also interchanges the α-string through β and the β-string through α. Thus r(α, β) = r(β, α). 

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https://doi.org/10.1090//ulect/066/03

CHAPTER 2

A basis for U In this section, let L be a Lie algebra over a field k and U an associative algebra over k. We say that ϕ : L → U is a homomorphism if: (1) ϕ is linear. (2) ϕ[X, Y ] = ϕ(X)ϕ(Y ) − ϕ(Y )ϕ(X) for all X, Y ∈ L. In other words, ϕ is a homomorphism of Lie algebras. A universal enveloping algebra of a Lie algebra L is a couple (U, ϕ) such that: (1) U is an associative algebra with 1. (2) ϕ is a homomorphism of L into U, so that ϕ[X, Y ] = [ϕ(X), ϕ(Y )] for all X, Y ∈ L. (3) If (A, ψ) is any other such couple then there exists a unique homomorphism θ : U → A such that θ ◦ ϕ = ψ and θ(1) = 1. For the existence and uniqueness2 of (U, ϕ) see, e.g., Jacobson [35] or Humphreys [31]. Theorem (Birkhoff-Witt3 Theorem). Let L be a Lie algebra over a field k and (U, ϕ) its universal enveloping algebra. Then: (a) ϕ is injective. (b) If L is identified with its image in U and if X1 , . . . , Xr is a linear basis for L, the monomials X1k1 X2k2 · · · Xrkr form a basis for U (where the ki are nonnegative integers). The proof here too can be found in [35] or [31]. Theorem 2. Again, let C be the base field. Assume that the basis elements {Hi , Xα } of L are as in Theorem 1 and are arranged in some order. For each choice ni , mα ∈ Z+ (i = 1, 2, . . . , l; α ∈ Σ) form the product, in U, of Hi of numbers mα all ni and Xα /mα ! according to the given order.∗ The resulting collection is a basis for the Z-algebra UZ generated by all Xαm /m! (m ∈ Z+ ; α ∈ Σ). Remark. The collection is a C-basis for U by the Birkhoff-Witt Theorem. Remark. UZ is called Kostant’s Z-form, see [38]. The proof of Theorem 2 will depend on a sequence of lemmas. 2 Up

to an isomorphism. ()

3 Sometimes

this result is called the Poincar´e-Birkhoff-Witt Theorem. ()

∗ Here

=

H  i ni

1 (Hi )(Hi ni !

− 1) · · · (Hi − (ni − 1)).

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2. A BASIS FOR U

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Lemma 4. Every polynomial over C in l variables H1 , . . . , Hl which takes on integral values at all integral values of the variables is an integral combination of the polynomials å l Ç  Hi i=1

ni

where ni ∈ Z+ and ni ≤ degree of the polynomial in Hi (and conversely, of course). Proof. Let f be such a polynomial. We may write Ç å r  Hl f= fj j j=0 each fj being a polynomial in H1 , . . . , Hl−1 . We replace Hl by Hl + 1 and take the difference. If we do this r times we get fr . Assuming the lemma true for polynomials in l − 1 variables (it clearly holds   for polynomials in no variables), hence for fr , we may subtract the term fr Hrl from f and complete the proof by induction on r.  Lemma 5. If α is a root and we write X, Y, H for Xα , X−α , Hα , then Å

Xm m!

ãÅ

Yn n!

min(m,n) Å

ã =

 j=0

Y n−j (n − j)!

åÅ ãÇ ã H − m − n + 2j X m−j . j (m − j)!

Proof. The case m = n = 1, XY = Y X + H, together with induction on n yield X(Y n /n!) = (Y n /n!)X + (Y n−1 /(n − 1)!)(H − n + 1). This equation and induction on m yield the lemma.  Hα  Corollary. Each n is in UZ . Proof. Set m = n in Lemma 5, write the right side as Ç å Ç å n−1 Å  Y n−j ã H − 2n + 2j Å X n−j ã H + (n − j)! j (n − j)! n j=0 then use induction on n and Lemma 4.



Lemma 6. Let LZ be the Z-span of the basis {Hi , Xα } of L. Then under the 1 Xαm preserves LZ , and the same adjoint representation, extended to U, every m! holds for LZ ⊗ LZ ⊗ · · · , any number of factors. Proof. Making

1 m m! Xα

act on the basis of LZ we get

1 m 1 X · Xβ = ±(r + 1)(r + 2) · · · (r + m − 1) Xβ+mα m! α m! if β = −α (see the definition of r = r(α, β) in Theorem 1), Xα · X−α = Hα , (Xα2 /2) · X−α = −Xα , Xα · Hi = − α, αi  Xα , and 0 in all other cases, which proves LZ is preserved. The second part follows by induction on the number of factors and the following lemma.  Lemma 7. Let U and V be L-modules and A and B additive subgroups thereof. If A and B are preserved by every Xαm /m! then so is A ⊗ B (in U ⊗ V ). Proof. Since X acts on U ⊗ V as X ⊗ 1 + 1 ⊗ X it follows from the binomial X j /j! ⊗ X m−j /(m − j)!, whence the lemma.  expansion that X m /m! acts as Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

2. A BASIS FOR U

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Lemma 8. Let S be a set of roots such that (a) α ∈ S ⇒ −α ∈ / S and (b) α, β ∈ S, α + β ∈ Σ ⇒ α + β ∈ S (e.g., the set of positive roots), arranged in some order. Then    1 mα X : mα ≥ 0 mα ! α α∈S

is a basis for the Z-algebra A generated by all Xαm /m! (α ∈ S; m ≥ 0). Proof. By the Birkhoff-Witt Theorem applied to the Lie algebra for which of the {Xα : α ∈ S} is a basis we see that every A ∈ A is a complex combination given elements. We must show all coefficients are integers. Write A = c Xαmα /mα ! + terms of at most the same total degree. We make A act on L ⊗ L ⊗ · · · (Σmα

⊗mα copies) and look for the component4 of A· in H ⊗H ⊗· · · . Any term of α X−α A other than the first leads to a zero component since there are either not enough factors (at least one is needed for each X−α ) or barely enough but with the wrong distribution (since Xβ · X−α is a nonzero element of H only if β = α), while the first leads to a nonzero component only if the Xα ’s and the X−α ’s are matched up, in all possible permutations. It follows that the component sought is cHα ⊗ · · · ⊗ Hα . Now each Hα is a primitive element of LZ (to see this imbed α in a simple system of roots and then use Lemma 1). Since A preserves LZ ⊗ LZ ⊗ · · · by Lemma 6 it follows that c ∈ Z, whence Lemma 8.  Hi −k Any formal product of elements of U of the form or Xαm /m! (m, n ∈ n + Z ; k ∈ Z) will be called a monomial and the total degree of the X’s its degree. Lemma 9. If β, γ ∈ Σ and m, n ∈ Z+ , then (Xγm /m!)(Xβn /n!) is an integral combination of (Xβn /n!)(Xγm /m!) and monomials of lower degree. Proof. This holds if β = γ obviously and if β = −γ by Lemma 5. Assume β = ±γ. By Lemma 8 applied to the set S of roots of the form iγ + jβ (i, j ∈ Z+ ), arranged in the order β, γ, β + γ, . . . we see that (Xγm /m!)(Xβn /n!) is an integral d combination of terms of the form (Xβb /b!)(Xγc /c!)(Xβ+γ /d!) · · · . The map Xα → α (α ∈ S) leads to a grading of the algebra A with values in the additive group generated by S. The left side of the preceding equation has degree nβ + mγ. Hence so does each term on the right, whence b, c, · · · are restricted by the condition bβ + cγ + d(β + γ) + · · · = nβ + mγ, hence also by b + c + 2d + · · · = n + m. Clearly b + c + d + · · · , the ordinary degree of the above term, can be as large as n + m only if b + c = n + m and d = · · · = 0 by the last condition, and then b = n and c = m by the first, which proves Lemma 9.  Lemma 10. If α and β are roots and f is any polynomial, then Xαn f (Hβ ) = f (Hβ − nα(Hβ )) Xαn . Proof. By linearity this need only be proved when f is a power of Hβ and then it easily follows by induction on the two exponents starting with the equation  Xα Hβ = (Hβ − α(Hβ )) Xα . Observe that α(Hβ ) is an integer. We can now prove Theorem 2. 4 It is assumed that the space L ⊗ L ⊗ . . . is represented as a direct sum of subspaces of the form Lγ1 ⊗ Lγ2 ⊗ . . ., where either γi is a root or γi = 0 (and then Lγi = H). () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

2. A BASIS FOR U

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  Proof of Theorem 2. By the corollary to Lemma 5 each Hni is in UZ , hence so is each of the proposed basis elements. We must show that each element of UZ is an integral combination of the latter elements, and for this it suffices to show that each monomial is. Any monomial may, by induction on the degree, Lemma 9 and Lemma 10, be expressed as an integral linear combination of monomials such that for each α the Xα terms all come together and in the order of the roots prescribed by Theorem 2, then also such that each α is represented at most once, because (X m /m!)(X n /n!) = m+n X m+n /(m + n)!. The H terms may now be n brought to the front (see Lemma 10), the resulting polynomial expressed as an Hi  integral combination of ’s by Lemma 4, each Hi term shifted to the position ni prescribed by Theorem 2, and Lemma 4 used for each Hi separately, to yield finally an integral combination of basis elements, as required.  Let L be a semisimple Lie algebra having Cartan subalgebra H. Let V be a representation space for L. We call a vector v ∈ V a weight vector if there is a linear function λ on H such that Hv = λ(H)v for all H ∈ H. If such a v = 0 exists, we call the corresponding λ a weight of the representation. Lemma 11. If v is a weight vector belonging to the weight λ, then for α a root we have Xα v is a weight vector belonging to the weight λ + α, if Xα v = 0. Proof. If H ∈ H, then HXα v = Xα (H + α(H))v = (λ + α)(H)Xα v.



Theorem 3. If L is a semisimple Lie algebra over C having Cartan subalgebra H, then (a) Every finite-dimensional irreducible L-module V contains a nonzero vector v + such that v + is a weight vector belonging to some weight λ and Xα v + = 0 (α > 0). (b) It then follows that if Vλ is the subspace of V consisting of weight vectors belonging to λ, then dim Vλ = 1. Moreoever, every weight μ has the form λ − Σα, where the α’s are positive roots. Also, V = ΣVμ (μ a weight). (c) The weight λ and the line containing v + are uniquely determined. (d) λ(Hα ) ∈ Z+ for α > 0. (e) Given any linear function λ satisfying (d), then there is a unique finitedimensional L-module V in which λ is realized as in (a). Proof. (a) There exists at least one weight on V since H acts as an abelian set of endomorphisms. We introduce a partial order on the weights by μ < ν if  ν − μ = α ( a sum of positive roots). Since the weights are finite in number, we have a maximal weight λ. Let v + be a nonzero weight vector belonging to λ. Since λ + α is not a weight for α > 0, we have  by Lemma 11 that Xα v + = 0 (α > 0). + (b) and (c). Now let W = Cv + μ 0). Let U − , U ◦ and U + be + the Birkhoffthe universal enveloping algebras of L− , H and © L respectively. ¶By © ¶ l m(α) ni ◦ , U , U+ Witt theorem, U − has a basis X has a basis α i=1 Hi α0 Xα ¶ © m(α) l p(α) ni basis where m(α), ni , p(α) ∈ Z+ . Hence i=1 Hi α0 Xα − ◦ + − U = U U U . Now W is invariant under U . Also, V = Uv + = U − U ◦ v + = U − v + Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

2. A BASIS FOR U

15

since V is irreducible, U + v + = 0, and U ◦ v + = Cv + . Hence V = W and (b) and (c) follow. (d) Hα is in the 3-dimensional subalgebra generated by Hα , Xα , X−α . Hence, by the theory of representations of this subalgebra, λ(Hα ) ∈ Z+ . (See Jacobson [35, pp. 82–85].) (e) See Cartier [12].  Corollary. If μ is a weight and α a root, then μ(Hα ) ∈ Z. Proof. This follows from (b) and (d) of Theorem 3 and the fact that β(Hα ) = β, α ∈ Z for α, β ∈ Σ.  Remark. λ and v + are called the highest weight and a highest weight vector respectively. By Theorem 2, we know that the Z-algebra UZ generated by Xαm /m! (α ∈ Σ, m ∈ Z+ ) has a Z-basis   å l Ç p(α)  Xαm(α)  Hi  X α + : m(α), ni , p(α) ∈ Z . m(α)! i=1 ni α>0 p(α)! α 0), and H ni respectively, then UZ = UZ UZ UZ . Lemma 12. If u ∈ UZ and v + is a highest weight vector, then the component of uv + in Cv + is nv + for some n ∈ Z.  Proof. We know that UZ+ v + = 0 and UZ− v + ⊆ μ0

where (A, B) = ABA−1 B −1 , where the product on the right is taken over all roots iα + jβ (i, j ∈ Z+ ) arranged in some fixed but arbitrary order, and where the cij s are unique integers depending on α, β, and the chosen ordering, but not on t or u. Furthermore, c11 = Nα,β . Observe that this can also be written as (Xα (t), Xβ (u)) = i j i,j>0 Xiα+jβ (ci,j t u ). Proof. In U[[t, u]] set f (t, u) = (exp tXα , exp uXβ )



exp(−cij ti uj Xiα+jβ )

where cij ∈ C. We shall show that we may choose the cij s in Z such that f (t, u) = 1. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 19

20

3. THE CHEVALLEY GROUPS

d d We note that t dt (exp tXα ) = tXα exp tXα . Then, we apply t dt to f (t, u), and using the product rule for differentiation we get d t f (t, u) = tXα f (t, u) dt + exp(tXα ) exp(uXβ )(−tXα ) exp(−tXα ) exp(−uXβ )  · exp(−cij ti uj Xiα+jβ )   + (exp tXα , exp uXβ ) exp(−cij ti uj Xiα+jβ ) · (−ckl ktk ul Xkα+lβ )

×



iα+jβ >kα+lβ

exp(−cij ti uj Xiα+jβ ).

iα+jβ ≤kα+lβ

We bring the terms −tXα and (∗)

− ckl ktk ul Xkα+lβ

to the front using, e.g., the relations (exp uXβ )(−tXα ) = (exp ad uXβ )(−tXα ) exp uXβ (see Lemma 14) and (exp ad uXβ )(−tXα ) = −tXα − Nβ,α tuXα+β − · · · We get an expression of the form Af (t, u) with A ∈ L[[t, u]]. Because f (t, u) is homogeneous of degree 0 relative to the grading6 t → −α, u → −β, Xγ → γ, A is also, and from formulas such as those above we see that ckl is involved in the term (∗) but otherwise only in terms of degree > k + l in t and u. Thus A =  k l k,l≥1 (−kckl +pkl )t u Xkα+lβ with pkl a polynomial in cij ’s for which i+j < k +l. Now we may inductively determine values of ckl ∈ C using the lexicographic ord f (t, u) = 0 implies f (t, u) = f (0, u) = 1. dering of the cij ’s such that A = 0. Then t dt To show that the cij s are integers, we examine the coefficient of ti uj in the definition of f (t, u). This coefficient is −cij Xiα+jβ +(terms coming from exponentials of multiples of Xkα+lβ with k + l < i + j). Using induction on i + j, we see that cij Xiα+jβ ∈ UZ . Hence, cij ∈ Z, by Theorem 2. If i = j = 1, the coefficient is  −c11 Xα+β + Nα,β Xα+β , so that c11 = Nα,β . Examples. (a) If α+β is not a root, the right side of the formula in Lemma 15 is 1, i.e., Xα and Xβ commute elementwise. In particular, if α and β are distinct simple roots, then Xα and X−β commute elementwise. (b) If α+β is the only root of the form iα+jβ, the right side is exp Nα,β tuXα+β , and Nα,β = ±(r + 1) with r = r(α, β) as in Theorem 1. (c) If all the roots have one length, the right side is 1 in case (a) and exp(±tuXα+β ) in case (b). Exercise. Work out for B2 and G2 . Corollary. If exp tXα , etc. in the formula in Lemma 15 are replaced by xα (t), etc., then the resulting equation holds for all t, u ∈ k. 6 On

L[[t, u]] we use the grading with values in K∗ . ()

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3. THE CHEVALLEY GROUPS

21

We call a set S of roots closed if α, β ∈ S, α + β ∈ Σ implies α + β ∈ S. The following are examples of closed sets of roots: (a) P = set of all positive roots. (b) P − {α}, α a simple root. (c) Pr = {α : ht α ≥ r, r ≥ 1}. We shall call a subset I of a closed set S an ideal if α ∈ I, β ∈ S, α + β ∈ S implies α + β ∈ I. We see that (a), (b) and (c) above are ideals in P . Lemma 16. Let I be an ideal in the closed set S. Let XS and XI denote the groups generated by all Xα (α ∈ S and α ∈ I, respectively). If α ∈ S implies −α ∈ / S, then XI is a normal subgroup of XS . Proof. This follows immediately from Lemma 15.



Lemma 17. Let S be a closed set of roots such / S, then that α ∈ S implies −α ∈ every element of XS can be written uniquely as α∈S xα (tα ) where tα ∈ k and the product is taken in any fixed order. Proof. We shall first prove the lemma in the case in which the ordering is consistent with heights, i.e., ht α < ht β implies α < β. S ⊂ P and ht is defined with respect to P . If α1 is the first element of S, then S − {α1 } is an ideal in S. Hence XS = Xα1 XS−{α1 } . Using induction on the size of S, we see XS = Xα . Now suppose y ∈ XS , y = xα (tα ). Since Xαk1 = 0, there is a weight vector v ∈ M corresponding to a weight λ such that Xα1 v = 0. Now yv = v + tα1 Xα1 v + z where v ∈ Vλ , tα1 Xα1 v ∈ Vλ+α1 , and z is a sum of terms from other weight spaces. Hence tα1 ∈ k is uniquely determined by y. Since xα1 (tα1 )−1 y ∈ XS−{α1 } , we may complete the proof of this case by induction.  The proof of Lemma 17 for an arbitrary ordering follows immediately from: Lemma 18. Let X be a group with subgroups X1 , X2 , . . . , Xr such that: (a) X = X1 X2 · · · Xr , with uniqueness of expression. (b) Xi Xi+1 · · · Xr is a normal subgroup of X for i = 1, 2, · · · , r. If p is any permutation of 1, 2, . . . , r, then X = Xp1 Xp2 · · · Xpr with uniqueness of expression. Proof. (Exercise) Consider X/Xr and use induction.



Corollary 1. The map t → xα (t) is an isomorphism of the additive group of k onto Xα . Corollary 2. Let P be the set of all positive roots and let U = XP . Then U = Xα with uniqueness of expression, where the product is taken over all α ∈ P arranged in any fixed order. Corollary 3. U is unipotent7 and is superdiagonal relative to an appropriate choice of basis for V k . Similarly, U − = X−P is unipotent and subdiagonal relative to the same choice of basis. Proof. Choose a basis of weight vectors and order them in a manner consistent with the partial ordering of the weights: μ precedes ν if μ − ν is a sum of positive roots.  7 A unipotent subgroup is a subgroup of a matrix group consisting of unipotent matrices, i.e., of matrices with all eigenvalues equal to 1. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Corollary 4. If i ≥ 1 let Ui be the group generated by all Xα with ht α ≥ i. We have then: (a) Ui is normal in U . (b) (U, Ui ) ⊆ Ui+1 , in particular, (U, U ) ⊆ U2 . (c) U is nilpotent. Corollary 5. If P = Q ∪ R with Q and R closed sets such that Q ∩ R = ∅, then U = XQ XR and XQ ∩XR = 1 (e.g., if α is a simple root, one can take Q = {α} and R = P − {α}). Example. If L = sll+1 , we have seen that the roots correspond to pairs (i, j) i = j, the positive roots to pairs (i, j) i < j, and that we may take Xij = Eij , the usual matrix unit. Thus, xij (t) = 1 + tEij . We see that U = {all unipotent, superdiagonal matrices}, U − = {all unipotent, subdiagonal matrices}, and that G is SLl+1 , the group of l + 1 square matrices of determinant 1. The nontrivial commutator relations are: (xij (t), xjk (u)) = xik (tu) if i, j, k are distinct. Lemma 19. For any root α and any t ∈ k∗ define wα (t) = xα (t)x−α (−t−1 )xα (t) and hα (t) = wα (t)wα (1)−1 . Then wα (t)−1 = wα (−t), and, (a) wα (t)Xβ wα (t)−1 = ct− β,α Xwα β where c = c(α, β) = ±1 is independent of t, k and the representation chosen, and c(α, β) = c(α, −β). (b) If v ∈ Vμk there exists v  ∈ Vwkα μ independent of t such that wα (t)v = t− μ,α v  . (c) hα (t) acts “diagonally” on Vμk as multiplication by t μ,α . Hence, H, the group generated by all hα (t) for all roots α and all t ∈ k, is abelian. (Note that wα is being used to denote both the defined automorphism and the reflection in the hyperplane orthogonal to α.) Proof. We prove this assuming k = C. The transfer of coefficients to an arbitrary field is almost immediate. We first show that wα (t)Hwα (t)−1 = wα H for all H ∈ H. By linearity it suffices to prove this for Hα , for if α(H) = 0 then Xα commutes with H so that both sides equal H. If H = Hα , the left side, because of Lemma 2 and the definitions of xα (t) and wα (t), is an element of the 3-dimensional algebra Xα , Yα , Hα  whose value depends on calculations within this algebra, not on the representation chosen. Taking the usual representation in sl2 , we get òï ò ï ò ï ò ï òï 1 t 0 t 1 0 1 t 1 0 Hα = = and wα (t) = 0 1 −t−1 0 0 −1 0 1 −t−1 1 so that the desired equation follows. We next prove (b). From  the i definitions of xα (t) and wα (t) it follows that if v  = wα (t)v, then v  = ∞ i=−∞ t vi where vi ∈ Vμ+iα (the sum is actually finite since there are only finitely many weights). Then for H ∈ H, Hv  = Hwα (t)v = wα (t)wα (t)−1 Hwα (t)v = wα (t)(wα H)v = μ(wα H)v  = (wα μ)(H)v  . Hence v  corresponds to the weight wα μ = μ − μ, α α. Thus the only nonzero term in the sum occurs for i = − μ, α. By (b) applied to the adjoint representation wα (t)Xβ wα (t)−1 = ct− β,α Xwα β where c ∈ C and is independent of t and of the representation chosen. Now wα (1) Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

3. THE CHEVALLEY GROUPS

23

is an automorphism of LZ and Xγ is a primitive element of LZ for all γ so c = ±1. Finally Hwα β = wα (1)Hβ wα (1)−1 = [wα (1)Xβ wα (1)−1 , wα (1)X−β wα (1)−1 ] = c(α, β)c(α, −β)Hwα β so c(α, β) = c(α, −β), which proves (a). Note that wα (t)−1 = wα (−t) so that hα (t) = wα (−t)−1 wα (−1). By (b) wα (−t)v = (−t)− μ,α v  and wα (−1)v = (−1)− μ,α v  . Hence wα (−t)−1 wα (−1)v = t μ,α v, proving (c).  Lemma 20. Write ωα for wα (1). Then: (a) ωα hβ (t)ωα−1 = hwα β (t) = an expression as a product of h’s, independent of the representation space. (b) ωα xβ (t)ωα−1 = xwα β (ct) with c as in Lemma 19(a). (c) hα (t)xβ (u)hα (t)−1 = xβ (t β,α u). Proof. To prove (a) we apply both sides to v ∈ Vμk . ωα hβ (t)ωα−1 v = ωα t wα μ,β ωα−1 v

(by Lemma 19(c) applied to ωα−1 v ∈ Vwkα μ )

= t wα μ,β v = t μ,wα β v = hwα β (t)v. By Lemma 19(c) ωα Xβ ωα−1 = cXwα β . Exponentiating this gives (b). By Lemma 19(c) applied to the adjoint representation hα (t)Xβ hα (t)−1 = t β,α Xβ . Exponentiating this gives (c).  Denote by (R) the following set of relations:8 (R1) xα (t)xα (u) = xα (t + u) (R2) (xα (t), xβ (u)) = xiα+jβ (cij ti uj ) (α+β = 0) with the cij as in Lemma 15. (R3) wα (t) = xα (t)x−α (−t−1 )xα (t) (R4) hα (t) = wα (t)wα (1)−1 (R5) ωα = wα (1) (R6) ωα hβ (t)ωα−1 = some expression as a product of h’s (independent of the representation space).9 (R7) ωα xβ (t)ωα−1 = xwα β (ct), c as in Lemma 19(a). (R8) hα (t)xβ (u)hα (t)−1 = xβ (t β,α u). Since all the relations in (R) are independent of the representation space chosen, results proved using only the relations (R) will be independent of the representation space chosen. Such results will be labeled (E) (usually for existence). Results proved using other information will be labeled (U) (usually for uniqueness). 8 (R1) follows from Corollary 1; (R2) from Lemma 15; (R3), (R4), and (R5) are definitions; (R6), (R7), and (R8) follow from Lemma 20. 9 This is a weak form of part (a) in Lemma 20. Relation (R6) is written in such a form that the results obtained with the help of this relation could be used in Chapter 6 (see Theorem 8(a)). () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Lemma 21. Let U be the group generated by all Xα (α > 0), H the group generated by all hα (t) and B the group generated by U and H. Then: (a) U is normal in B and B = U H. (E) (b) U ∩ H = 1. (U) Proof. Since conjugation by hα (t) preserves Xβ (by (R8)) U is normal in B and (a) holds. Relative to an appropriate basis of V any element of U ∩ H is both diagonal and unipotent, hence = 1.  Example. In SLn , H = {diagonal matrices}, U = {unipotent superdiagonal matrices}, B = {superdiagonal matrices}. Lemma 22. Let N be the group generated by all wα (t), H be the subgroup generated by all hα (t) and W the Weyl group. Then: (a) H is normal in N . (E) (b) There exists a homormorphism ϕ of W into N/H such that (E) ϕ(wα ) = Hwα (t) for all roots α. (c) ϕ is an isomorphism. (U) Proof. Since by (R6) conjugation by ωα preserves H and by (R4) and (R5) wα (t) = hα (t)ωα , (a) holds. Since Hwα (t) = Hwα (t)wα (1)−1 wα (1) = Hwα (1),

α = Hwα (t). Then since wα (1) ∈ w

α and Hwα (t) is independent of t. Write w

α , 1 = wα (1)wα (−1) ∈ w

α2 . Hence (∗) w

α2 = 1. Also ωβ = wβ (1) ∈ w

β wα (−1) ∈ w so ωα ωβ ωα−1 ∈ w

α w

β w

α−1 . But ωα ωβ ωα−1 = ωα xβ (1)x−β (−1)xβ (1)ωα−1

(by (R3))

= xwα β (c)x−wα β (−c)xwα β (c)

(by (R7))

= wwα β (c) ∈ w

wα β Thus (∗) w

α w

β w

α−1 = w

wα β . By Appendix IV.40 the relations (∗) form a defining set for W . Thus there exists a homomorphism ϕ : W → N/H such that ϕwα = w

α = Hwα (t). ϕ is clearly onto. Suppose w ∈ ker ϕ. If w = wα1 wα2 · · · , a product of reflections, then wα1 (1)wα2 (1) · · · = h ∈ H. Conjugating Xα by wα1 (1)wα2 (1) · · · we get Xwα and cojugating by h we get Xα . Hence Xwα = Xα for all roots α. Since wα = α for all α implies w = 1 the proof is completed by the following lemma.  Lemma 23. If α and β are distinct roots then Xα = Xβ . Proof. We know that Xα is nontrivial. If α and β have the same sign, the result follows from Lemma 17. If they have opposite sign, then one is superdiagonal unipotent, the other subdiagonal (relative to an appropriate basis), and the result again follows.  Convention. If n ∈ N represents w ∈ W (under ϕ : W → N/H) we will write wB (Bw) in place of nB (Bn). Lemma 24. If α is a simple root then Bwα B = Bwα Xα and B ∪ Bwα B is a group. (E) Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

3. THE CHEVALLEY GROUPS

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Proof. It is enough to prove this with B replaced by each of H, UP −{α} , and Uα because B is the product of these groups (see Corollary 5 of Lemma 18). For the first two cases this is so because wα normalizes these groups, which holds in  the second case because the relection wα permutes the elements of P − {α}. Lemma 25. If w ∈ W and α is a simple root, then: (a) If wα > 0 (i.e., if N (wwα ) = N (w) + 1 (see Appendix II.17)) then BwB · Bwα B ⊆ Bwwα B. (b) In any case BwB · Bwα B ⊆ Bwwα B ∪ BwB.

(E) (E)

Proof. (a) BwB · Bwα B = BwXα XP −{α} Hwα B = BwXα w−1 wwα wα−1 XP −{α} wα wα−1 Hwα B = Bwwα B (for wXα w−1 ⊆ B, wα−1 XP −{α} wα ⊆ B and wα−1 Hwα ⊆ B). (b) If wα > 0, (a) gives the result. If wα < 0 set w = wwα . Then w α > 0 and w = w wα . By (a), BwB · Bwα B = Bw wα B · Bwα B ⊆ Bw B · Bwα B · Bwα B ⊆ Bw B(B ∪ Bwα B) 



⊆ Bw B ∪ Bw wα B

(by Lemma 24) (by (a))

= BwB ∪ Bwwα B  Corollary. If w ∈ W and w = wα wβ · · · is an expression of minimal length of w as a product of simple reflections then BwB = Bwα B · Bwβ B · · · . Lemma 26. Let G be the Chevalley group G = Xα |all α a root. Then G is (E) generated by all Xα , ωα for α a simple root. Proof. We have ωα Xβ ωα−1 = Xwα β . Since the simple reflections generate W and every root is conjugate under W to a simple root the result follows.  Theorem 4 (Bruhat, Chevalley).  (a) w∈W BwB = G. (E)   (b) BwB = Bw B ⇒ w = w . (U) Thus any system of representatives for N/H is also a system of representatives for B\G/B.   Proof. (a) By Lemma 26 w∈W BwB contains a set of generators for G. Since w∈W BwB is closed under multiplication by these generators (by Lemma 25) and reciprocation it is equal to G. (b) Suppose BwB = Bw B with w, w ∈ W . We will show by induction on N (w) that w = w . (Here N (w) is as in the Appendix II.) If N (w) = 0 then w = 1 so w ∈ B. Then w Bw−1 = B so w P = P and w = 1 (see Appendix II.23). Assume N (w) > 0 and choose α simple so that N (wwα ) < N (w). Then wwα ∈ Bw BBwα B ⊆ Bw B ∪ Bw wα B = BwB ∪ Bw wα B. Hence by induction Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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wwα = w or wwα = w wα . But wwα = w implies wα = 1 which is impossible.  Hence wwα = w wα so w = w . Remark. The groups B, N form a BN pair in the sense of J. Tits [63]. We shall not axiomatize this concept but adapt certain arguments, such as the last one, to the present context. Theorem 4 . For a fixed w ∈ W choose ωw representing w in N . Set Q = P ∩ w−1 (−P ), R = P ∩ w−1 P (as before P denotes the set of positive roots). Write Uw for XQ . Then: (E) (a) BwB = Bωw Uw . (b) Every element of BwB has a unique expression of this form. (U) Proof. (a) BwB = BwXR XQ H = BwXR w

−1

= BwXQ H

(by Lemma 17 and Lemma 21)

wXQ H (since wXR w−1 ⊆ B)

= Bωw XQ . −1 (b) If bωw x = b ωw x then b−1 b = ωw xx−1 ωw . Relative to an appropriate basis this is both superdiagonal and subdiagonal unipotent and hence = 1. Thus  b = b and x = x .

Exercise. (a) Prove B is the normalizer in G of U and also of B. (b) Prove N is the normalizer in G of H if k has more than 3 elements. Examples. (a) Let L = sln so that G = SLn , and B, H, N are respectively the superdiagonal, diagonal, monomial10 subgroups, and W may be identified with the group of permutations of the coordinates. Going to G = GLn for convenience, we get from Theorem 4: (∗) the permutation matrices Sn form a system of representatives for B\G/B. We shall give a simple direct proof of this. Here k can be any division ring. Given x ∈ G, choose b ∈ B to maximize the total number of zeros at the beginnings of all of the rows of bx. These beginnings must all be of different lengths since otherwise we could subtract a multiple of some row from an earlier row, i.e., modify b, and increase the total number of zeros. It follows that for some w ∈ Sn , wbx is superdiagonal, whence x ∈ Bw−1 B. Now assume BwB = Bw B with w, w ∈ Sn . Then w−1 bw is superdiagonal for some b ∈ B. Since w, w are permutation matrices and the matrix positions where the identity is nonzero are included among those of b, we conclude that w−1 w is superdiagonal, whence w = w , which proves (∗). Next we will give a geometric interpretation of the result just proved. Let V be the underlying vector space. A flag in V is an increasing sequence of subspaces V1 ⊂ V2 ⊂ · · · ⊂ Vn where dim Vi = i. Associated with the chosen basis {v1 , . . . , vn } of V there is a flag F1 ⊂ · · · ⊂ Fn defined by Fi = v1 , . . . , vi  called the standard flag. Now G acts on V and hence on flags. B is the stabilizer of the standard flag, so B\G/B is in one-to-one correspondence with the set of G orbits of pairs of flags. 10 I.e., the subgroup of all matrices that have exactly one nonzero element in each row and in each column. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

3. THE CHEVALLEY GROUPS

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Define a simplex to be a set of points {p1 , . . . , pn } of V such that dim p1 , . . . , pn  = n. A flag V1 ⊂ · · · ⊂ Vn is said to be incident with this simplex if Vi = pπ1 , . . . , pπi  for some π ∈ Sn . Hence there are n! flags incident with a given simplex. It can be shown, by induction on n (see Steinberg [57]), that (∗) given any two flags there is a simplex incident with both. Thus associated to each pair of flags there is an element of Sn , the permutation which transforms one to the other. Hence B\G/B corresponds to Sn . Thus (∗) is the geometric interpretation of the Bruhat decomposition for sln . (b) Consider L = {X ∈ slm : XA + AX t = 0} where A is fixed and nonsingular (i.e., consider the invariants of the automorphism X → A(−X t )A−1 ). If m = 2n and A is skew this gives an algebra of type Cn , if m = 2n and A is symmetric this gives an algebra of type Dn , and if m = 2n + 1 and A is symmetric this gives an algebra of type Bn . If we take ⎤ ⎡ 1 ⎥ ⎢ . ⎥ ⎢ .. ⎥ ⎢ ⎥ ⎢ 1 ⎥ ⎢ A=⎢ ⎥ −1 ⎥ ⎢ ⎥ ⎢ . . ⎦ ⎣ . −1 in the first case and



1

⎢ A=⎣

..

.

⎤ ⎥ ⎦

1 in the second  and third cases an element y ∈ L is superdiagonal if and only if ad y preserves α>0 Lα (with the usual ordering of roots). Exponentiating we get the σ invariants of X → A(X −1 )t A−1 (that is XAX t = A). In the first case we get G ⊆ Spm , in the second and third cases G ⊆ SOm (relative to a form of maximal index). (For the proof that equality holds see Ree [46].) The automorphism σ above preserves the basic ingredients B, H, N of the Bruhat decomposition of SLm . From this a Bruhat decomposition for Spm and SOm can be inferred. By a slight modification of the procedure, we can at the same time take care of unitary groups. (c) If L is of type G2 it is the derivation algebra of a split Cayley algebra. The corresponding group G is the group of automorphisms of this algebra. Since the results labeled (E) in Theorem 4 depend only on the relations (R) (which are independent of the representation chosen) we may extract from the discussion the following result. Proposition. Let G be a group generated by elements labeled xα (t) (α ∈ Σ, t ∈ k) such that the relations (R) hold and let U  , H  , . . . be defined in G as U, H, . . . are in G. (1) Every element of U  can be written in the form α>0 xα (tα ). (2) For each w ∈ W , write w = wα wβ · · · a product of reflections. Define  = ωα ωβ · · · (where ωα = wα (1)). Then every element of G can be ωw   written u h ωw v (where u ∈ U  , h ∈ H  , v  ∈ Uw ). Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Corollary 1. Suppose G is as above and ϕ is a homomorphism of G onto G such that ϕ(xα (t)) = xα (t) for all α and t. Then: (a) Uniqueness of expression holds in (1) and (2) above. (b) ker ϕ ⊆ Z(G ) ⊆ H  , where Z(G ) denotes the center of G .   xα ( tα ). Applying ϕ we get xα (tα ) = Proof. (a) Suppose xα (tα ) = tα ) and by Lemma 17 tα =  tα for all α. Hence ϕ|U  is an isomorphism. Now if xα (     v =u  v , by applying ϕ we get ϕ(u )ϕ(h )ωw ϕ(v  ) = ϕ( u )ϕ( h )ωw ϕ( v  ). u h ωw h ωw        By Theorem 4 and Lemma 21 ϕ(u ) = ϕ( u ) and ϕ(v ) = ϕ( v ). Hence u = u  and   = h ωw so h =  h . v  = v so h ωw   (b) Let x = u h ωw v ∈ ker ϕ. Then 1 = ϕ(u )ϕ(h )ωw ϕ(v  ) ∈ U Hωw Uw ;   ) = 1. Hence u = v  = 1 so x = h = so w = 1, ωw = 1, ϕ(u ) = 1, ϕ(v Ä ä  β,α hα (tα ). Then x xβ (u)(x )−1 = xβ u by (R8). Applying ϕ we see that α tα β,α   = 1. Hence x commutes with xβ (u) for all β and u, so is in the center α tα of G . To complete the proof it is enough to show that Z(G) ⊂ H (for we have shown ker ϕ ⊂ H  ). If x = uhωw v ∈ Z(G) and w = 1 then there exists α > 0 such that wα < 0. Then xxα (1) = xα (1)x which contradicts Theorem 4 . Hence w = 1 so x = uh. Let w0 be the element of W making all positive roots negative. −1 is both superdiagonal and subdiagonal. Since h is diagonal, u Then x = ωw0 xωw 0 is diagonal, and also unipotent. Hence u = 1 and x = h ∈ H. 

Corollary 2. Z(G) ⊆ H. Corollary 3. The relations (R) and those in H on the hα (t) form a defining set of relations for G. Proof. If the relations in H are imposed on H  then ϕ in Corollary 1 becomes an isomorphism by (b).  Corollary 4. If G is constructed as G from L, k, . . . but using perhaps a different representation space V  in place of V , then there exists a homomorphism ϕ of G onto G such that ϕ(xα (t)) = xα (t) if and only if there exists a homomorphism θ : H  → H such that θ(hα (t)) = hα (t) for all α and t. Proof. Clearly if ϕ exists then θ exists. Conversely assume θ exists. Matching up the generators of H  and H, we see that the relations in H  form a subset of those in H. By Corollary 3 and the fact that the relations (R) are the same for G and G, the relations on xα (t), . . . in G form a subset of those on xα (t), . . . in G. Thus ϕ exists.  So far the structure of H has played a minor role in the proceedings. To make the preceding results more precise we will now determine it. We recall that H is the group generated by all hα (t) (α ∈ Σ, t ∈ k) and that (∗)

hα (t) acts on the weight space Vμ as multiplication by t μ,α .

Also, we recall that by Theorem 3(e), a linear function μ on H is the highest weight of some irreducible representation provided μ, α = μ(Hα ) ∈ Z+ for all α > 0. Clearly, it suffices that μ, α ∈ Z+ for all simple roots αi . Define λi , i = 1, 2, . . . , l by λi , αj  = δij . We see that λi occurs as the highest weight of some irreducible representation, and we call λi a fundamental weight. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Lemma 27. (a) The additive group generated by all the weights of all representations forms a lattice L1 having {λi } as a basis. (b) The additive group generated by all roots is a sublattice L0 of L1 . Moreover, (αi , αj ) i, j = 1, 2, . . . , l is a relation matrix for L1 /L0 , which is thus finite. (c) The additive group generated by all weights of a faithful representation on V forms a lattice LV between L0 and L1 . Proof. Part (a) is immediate from the definition of the fundamental weights.  (b) cij λj (cij ∈ C) then αi , αk  = cik and  If αi is a simple root and αi = αi = αi , αj  λj . (c) If α is a root, then since Xα = 0 there exists 0 = v ∈ Vμ for some weight μ  with 0 = Xα v ∈ Vμ+α . Hence α = (μ + α) − μ ∈ LV and L0 ⊆ LV ⊆ L1 . Remark. All lattices between L0 and L1 can be realized as in Lemma 27(c) by an appropriate choice of V . In particular, LV = L0 if V corresponds to the adjoint representation, and LV = L1 if V corresponds to the sum of the representations having fundamental weights as dominant weights. Lemma 28 (Structure of H). (a) For each α, hα (t) is multiplicative as a function of t. (b) H is an abelian group generated by the hi (t)’s (with hi (t) = hαi (t)). μ,α (c) li=1 hi (ti ) = 1 if and only if li=1 ti i = 1 for all μ ∈ LV . © ¶ l β,αi l (d) Z(G) = = 1 for all β ∈ L0 , hence is finite. i=1 hi (ti ) : i=1 ti Proof. (a), (b), and (c) follow from (∗) above. (a) and (c) are immediate and  (b) results from t μ,α = tμ(Hα ) = tμ(Σ ni Hi ) = tΣ ni μ,αi if Hα = ni Hi . For (d), β,α we note that li=1 hi (ti ) commutes with xβ (u) if and only if li=1 ti i = 1 by Lemma 19(c).  Corollary. (a) If LV = L1 , then every h ∈ H can be written uniquely as h = li=1 hi (ti ), ti ∈ k ∗ . (b) If LV = L0 , then G has center 1. Corollary 5 (To Theorem 4 ). Let G be a Chevalley group as usual and let G be another Chevalley group constructed from the same L and k as G but using V  in place of V . If LV  ⊇ LV , then there exists a homomorphism ϕ : G → G such that ϕ(xα (t)) = xα (t) for all α, t and ker ϕ ⊆ Z(G ), where xα (t) corresponds to xα (t) in G . If LV = LV  , then ϕ is an isomorphism. 

Proof. There exists a homomorphism θ : H  → H such that θ(hi (t)) = hi (t) by Lemma 28(c). If α is any root and Hα = ni Hi , ni ∈ Z, then hα (t) = hi (t)ni and similarly for hα (t). Hence θ(hα (t)) = hα (t). By Corollary 4 to Theorem 4 , ϕ exists. By Corollary 1, ker ϕ ⊆ Z(G ). If LV = LV  , we have a homomorphism ψ : G → G such that ψ(xα (t)) = xα (t). Hence ψ ◦ ϕ = idG , ϕ ◦ ψ = idG , and ϕ is an isomorphism.  Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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We call the Chevalley groups G0 and G1 corresponding to the lattices L0 and L1 the adjoint group and the universal group 11 respectively. If G = GV is a Chevalley group corresponding to the lattice LV , then by Corollary 5, we have central homomorphisms α and β such that α : G1 → GV and β : GV → G0 . We call ker α the fundamental group of G and we see that ker β = Z(G). Exercise. The center of the fundamental group, i.e., the fundamental group of the adjoint group, is isomorphic to Hom(L1 /L0 , k∗ ). E.g., if k = C, the last group is isomorphic with L1 /L0 . Also in this case Z(GV ) ∼ = LV /L0 , and the fundamental group of GV is isomorphic to L1 /LV . In the following table, we list some information known about the lattices and Chevalley groups of the various Lie algebras L: Type of L L1 /L0 Al Zl+1 Bl Z2 Cl Z2 D2n+1 Z4 D2n Z2 × Z2 E6 Z3 E7 Z2 E8 Z1 F4 Z1 G2 Z1

G0 PSLl+1 PSO2l+1 = SO2l+1 PSp2l PSO4n+2 PSO4n

GV

SO4n+2 SO4n

G1 SLl+1 Spin2l+1 Sp2l Spin4n+2 Spin4n

G0 G0 G0

= = =

G1 G1 G1

Here GV is a Chevalley group other than G0 and G1 , Zn is the cyclic group of order n, SOn is the special orthogonal group, Spinn is the spin group, Spn is the symplectic group, and PG denotes the projective group of G. To obtain the column headed by L1 /L0 one reduces the relation matrix (αi , αj ) to diagonal form. To show, for example, that SLn is the universal group of L = sln of type An−1 , we let ωi be the weight ωi : diag(a1 , . . . , an ) → ai . Then if λi = ω1 + ω2 + · · · + ωi , 1 ≤ i ≤ n − 1, we have λi (Hj ) = λi (Ejj − Ej+1,j+1 ) = δij . Hence the fundamental weights are in the lattice associated with this representation. Since the center of SLn is generically cyclic of order n, it follows that L1 /L0 is isomorphic to Zn in this case. Exercise. If G is a Chevalley group, G1 , G2 , . . . , Gr subgroups of G corresponding to irreducible components of Σ, then: (a) Each Gi is normal in G and G = G1 G2 · · · Gr . (b) G is universal (respectively adjoint) if and only if each Gi is. (c) In each case in (b), the product in (a) is direct.

11 A

more standard name for this group is the simply-connected group. ()

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3. THE CHEVALLEY GROUPS

Corollary 6. If α Xα , X−α  such that ï 1 ϕα 0 ï 0 ϕα −1

31

is a root, then there exists a homomorphism ϕα : SL2 → t 1 1 0

ò

ï = xα (t),

ò = ωα ,

ϕα ï ϕα

1 0 t 1

t 0 0 t−1

ò = x−α (t), ò = hα (t).

Moreover, ker ϕα = {1} or {±1} so that Xα , X−α  is isomorphic to either SL2 or PSL2 . Proof. Let L1 be of rank 1 spanned by X, Y and H with [X, Y ] = H, [H, X] = 2X and [H, Y ] = −2Y . Now L1 has a representation ï ò ï ò ï ò 0 1 0 0 1 0 X → Y → H → 0 0 1 0 0 −1 as sl2 on a vector space V  and a representation X → Xα , Y → X−α , H → Hα on the same vector space V as the original representation of L. Since SL2 is universal, the required homomorphism ϕ exists and has ker ϕ ⊆ {±1} by Corollary 5.  Exercise. If G is universal, each ϕα is an isomorphism.

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https://doi.org/10.1090//ulect/066/05

CHAPTER 4

Simplicity of G The main purpose of this chapter is to prove the following theorem: Theorem 5 (Chevalley, Dickson). Let G be an adjoint group and assume L is simple (Σ indecomposable). If |k| = 2, assume L is not of type A1 , B2 , or G2 . If |k| = 3, assume L is not of type A1 . Then G is simple. Remark. The cases excluded in Theorem 5 must be excluded. If |k| = 2, then G has A3 , A6 , SU3 (3) as a normal subgroup of index 2 if L is of type A1 , B2 , G2 , respectively. If |k| = 3 and L is of type A1 , then A4 is a normal subgroup of G of index 2. Here A denotes the alternating group. A proof of Theorem 5 essentially due to Iwasawa and Tits will be given here in a sequence of lemmas. Lemma 29. Let G be a Chevalley group. If w ∈ W , w = wα wβ · · · is a minimal expression as a product of simple reflections, then wα , wβ , . . . ∈ G1 , the group generated by B and wBw−1 . Proof. We know w−1 α < 0 by the minimality of the expression (See Appendix II.19 and II.22). Hence if β = −w−1 α > 0, then G1 ⊇ wXβ w−1 = Xwβ = X−α . Thus wα ∈ G1 . Since wα wBw−1 wα−1 ⊆ G1 and since length wα w < length w, we may complete the proof by induction.  Lemma 30. If G is again any Chevalley group, if π is a subset of the  set of simple roots, if Wπ is the group generated by all wα , α ∈ π, and if Gπ = w∈Wπ BwB, then (a) Gπ is a group. (b) The 2l groups so obtained are all distinct. (c) Every subgroup of G containing B is equal to one of them. Proof. Part (a) follows from BwB · Bwα B ⊆ Bwwα B ∪ BwB.  (b) Suppose π, π  are distinct subsets  of the set of simple roots. Say α ∈ π , α∈ / π. Now wα α = −α and wα = α+ β∈π cβ β if w ∈ Wπ . Thus wα α = wα, since simple roots are linearly independent. Hence wα ∈ / Wπ , Wπ = Wπ , and Gπ = Gπ since distinct elements of the Weyl group correspond to distinct double cosets.

(c) Let A be any subgroup containing B. Set π = {α  : α simple, wα ∈ A}. We shall show A = Gπ . Clearly, A ⊇ Gπ . Since G = w∈W BwB and A ⊇ B, we need only show w ∈ A implies w ∈ Gπ to get A ⊆ Gπ . Let w ∈ A, w = wα wβ · · · , a minimal expression of w as a product of simple reflections. By  Lemma 29, wα , wβ , . . . ∈ A. Hence, α, β, . . . ∈ π, w ∈ Wπ , and w ∈ Gπ . Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 33

34

4. SIMPLICITY OF G

A group conjugate to some Gπ is called a parabolic subgroup of G. We state without proof some further properties of parabolic subgroups which follow from Lemma 29. (1) (2) (3) (4)

No two Gπ ’s are conjugate. Each parabolic subgroup is its own normalizer in G. Gπ ∩ Gπ = Gπ∩π . B ∪ BwB (w ∈ W ) is a group if and only if w = 1 or w is a simple reflection.

Example. If G = SLn , then π corresponds to a partition of the n × n matrices into blocks with the diagonal blocks being square matrices. Clearly there are 2n−1 possibilities for such partitions. Gπ is then the subset of SLn of matrices whose subdiagonal blocks are all zero. Lemma 31. Let L be simple and let G be the adjoint Chevalley group. If N = 1 is a normal subgroup of G, then N B = G. Proof. We first show N  B. Suppose N ⊆ B and 1 = x ∈ N , x = uh, u ∈ U , h ∈ H. If u = 1, then for some w ∈ W , wxw−1 ∈ / B, a contradiction. If u = 1, then h = 1. Since G is adjoint, it has center 1, and hxα (t)h−1 = xα (t ) with t = t for some t, t ∈ k, α ∈ Σ. Hence (h, xα (t)) = xα (t − t) ∈ N , xα (t − t) = 1 and we are back in the first case. We must now prove the lemma. By Lemma 30(c), N B = Gπ for some π. We must show π contains all simple roots. Suppose it does not. Since N  B, we see that π = ∅. Also since Σ is indecomposable, we can find simple roots α, β with α ∈ π, β ∈ / π and α not orthogonal to β. Let b1 wα b2 ∈ N , bi ∈ B, then b2 (b1 wα b2 )b−1 2 = bwα ∈ N with b = b2 b1 ∈ B. Then wβ bwα wβ−1 ∈ N ∩ (Bwα wβ B ∪ Bwβ wα wβ B) by Lemma 25(b). Hence either wα wβ ∈ Wπ or wβ wα wβ ∈ Wπ . Now wβ wα wβ = wγ where γ = wβ α = α − α, β β. Since α, β = 0, γ is not a simple root and N (wβ wα wβ ) = 1, so that N (wβ wα wβ ) ≥ 3 by Appendix II.20. Hence wα wβ and wα wβ wα are both expressions of minimal length. By Lemma 29, wβ ∈ Wπ , a  contradiction. Thus, π is the set of all simple roots and N B = Gπ = G. Lemma 32. If L and G are as in Theorem 5, then G = G , the derived group of G. Before proving Lemma 32, we first show that Theorem 5 follows from Lemmas 31 and 32. Let N = 1 be a normal subgroup of G. By Lemma 31, N B = G so G/N ∼ = B/(B ∩ N ). Now G/N equals its derived group and B/(B ∩ N ) is solvable. Hence G/N = 1 and N = G. Instead of proving Lemma 32 directly, we prove the following stronger statement: Lemma 32 . If L is as in Theorem 5 then G = G holds in any group G in which relations (R) hold, in fact in which the relations: (A) (xβ (t), xγ (u)) = xiβ+jγ (cij ti uj ) (B) hα (t)xα (u)hα (t)−1 = xα (t2 u) hold. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

4. SIMPLICITY OF G

35

Proof. Since G is generated by the Xα ’s we must show that every Xα ⊆ G . We will do this in several steps, excluding as we proceed cases already treated. The first step takes us almost all the way. (a) Assume |k| ≥ 4. We may choose t ∈ k∗ , t2 = 1. Then (hα (t), xα (u)) = xα ((t2 − 1)u). Since α and u are arbitrary, every Xα ⊆ G . By (a) we may henceforth assume that the rank l is at least 2 and that |k| = 2 or 3. By the corollary to Lemma 15, we may write the right side of (A) as xβ+γ (Nβ,γ tu) ·  , the factor with i = j = 1 having been isolated. We will use the fact (∗) Nβ,γ = ±(r + 1) with r = r(β, γ) as in Theorem 1, the maximum number of times one can subtract γ from β and still have a root. (b) Assume that α is a root which can be written β +γ so that no other positive integral combination of β and γ is a root and Nβ,γ = 0.12 Then Xα ⊆ G , as follows at once from (A) with  = 1. This covers the following cases: (1) If all roots have the same length: types Al , Dl , El . (2) Bl (l ≥ 3), α long; B2 , α long, |k| = 3. (3) Cl (l ≥ 3), α short; or α long and |k| = 3. (4) F4 . (5) G2 , α long. To see this we use the fact that all roots of the same length are congruent under the Weyl group, imbed α in an appropriate root system based on a pair of simple roots, and use (∗). In all cases but the second cases in (2) and (3) this system can be chosen of type A2 with β and γ roots of the same length as α, while in those cases it can be chosen of type B2 with α and γ short roots. Because of the exclusions in the theorem, this leaves the following cases: (6) Bl (l ≥ 2), α short. (7) G2 , α short, |k| = 3. (8) Cl (l ≥ 3), α long, |k| = 2. (c) If (6) or (7) holds, then Xα ⊆ G . In both of these cases we can find roots β, γ so that α = β + γ, all other roots iβ + jγ (i, j positive integers) are long, and Nβ,γ = 0: in (6) we can choose β long and γ short, in (7) both short. Then  belongs to G by cases already treated, hence so does Xα , by (A). (d) If (8) holds, then Xα ⊆ G . Choose roots β, γ with β long, γ short, and α = β + 2γ. Since Xβ+γ ⊆ G because β + γ is short, our assertion will follow from C12 = 0 in (A), hence from the next lemma.  Lemma 33. If β and γ form a simple system of type B2 with β long and γ short, then (xβ (t), xγ (u)) = xβ+γ (±tu)xβ+2γ (±tu2 ). Proof. By Lemma 14, we have xγ (u)Xβ xγ (u)−1 = exp (ad uXγ ) Xβ = Xβ + uNγ,β Xβ+γ + u2 12 Nγ,β Nγ,β+γ Xβ+2γ . 12 Here

we assume that Nβ,γ = 0 in the field k. ()

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36

4. SIMPLICITY OF G

Here Nγ,β = ±1 and Nγ,β+γ = ±2 since β − γ is not a root. If we multiply this equation by −t, exponentiate, observe that the three factors on the right side commute, and then shift the first of them to the left, we get Lemma 33.  The proof of Theorem 5 is now complete. In the course of this discussion, we have established the following result: Corollary. If Σ is indecomposable and of rank > 1 and if α is any root, then there exist roots β and γ and a positive integer n such that α = β + nγ and c1n = 0 in the relations (A) of Lemma 32  . Corollary (To Theorem 5). If |k| ≥ 4 and G is a Chevalley group based on k, then every solvable normal subgroup of G is central and hence finite. Proof. Since the center of a Chevalley group is always finite by Lemma 28(d), we need only prove the first statement. Also we may assume G = G0 , the adjoint group, since by Corollary 5 to Theorem 4 , there is a homomorphism ϕ of G onto G0 with ker ϕ ⊆ Z(G) and Z(G0 ) = 1. Now we may write G = G1 G2 · · · Gr where Gi (i = 1, 2, . . . , r) is the adjoint group corresponding to an indecomposable subsystem of Σ. By Theorem 5, each Gi is simple. Thus any normal subgroup of G is a product of some of the Gi ’s. If it is also solvable, the product is empty and the subgroup is 1. 

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https://doi.org/10.1090//ulect/066/06

CHAPTER 5

Chevalley Groups and Algebraic Groups The significance of the results so far to the theory of semisimple algebraic groups will now be indicated. Let k be an algebraically closed field. A subset V ⊆ kn is said to be algebraic if there exists a subset P ⊆ k[x1 , . . . , xn ], the algebra of polynomials on kn , such that V = {v = (v1 , . . . , vn ) ∈ kn : p(v1 , . . . , vn ) = 0 for all p ∈ P} . The algebraic subsets of kn are the closed subsets of the Zariski topology on kn . For V ⊆ kn set Ik (V ) = {p ∈ k[x1 , . . . , xn ] : p(v1 , . . . , vn ) = 0 for all (v1 , . . . , vn ) ∈ V }. Let r = n2 + 1. Define D(x) ∈ k[x0 ; xij ]1≤i,j≤n by D(x) = 1 − x0 det(xij ). Then GLn (k) = {v ∈ kr : D(v) = 0} is an algebraic subset of kr . G is a matrix algebraic group if G is a subgroup of GLn (k) for some n and some algebraically 2 closed field k, and G is an algebraic subset of kn +1 . If k0 is a subfield of k, G is defined over k0 if Ik (G) has a basis of polynomials with coefficients in k0 . Examples. (a) SLn (k) (b) Superdiagonal subgroup. (c) Diagonal subgroup. ßï ò™ 1 t (d) = Ga = additive group. 0 1 ßï ò™ t 0 (e) = Gm = multiplicative group. 0 t−1 (f) Sp2n (g) SOn (h) Any finite group. The groups in (a)–(e) are defined over the prime field.13 Whether Sp2n , SOn are or not depends on the coefficients of the defining bilinear forms. The groups in (h) are not connected in the Zariski topology, the others are. A map of algebraic groups ϕ : G → H is a homomorphism if it is a group homomorphism and each of the matrix coefficients ϕ(g)ij is a rational function of the gij . A homomorphism ϕ : G → H is an isomorphism if there exists a homomorphism ψ : H → G such that ϕψ = idH and ψϕ = idG . A homomorphism ϕ : G → H is defined over k0 if each of the rational functions above has coefficients in k0 . 13 I.e.,

over the field Q or over the field Fp where p is a prime number. ()

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38

5. CHEVALLEY GROUPS AND ALGEBRAIC GROUPS

The following results are proved in Lazard [40] (see also Springer [55]):14 (1) Let G be a matrix algebraic group. Then the following are equivalent: (a) G is connected (in the Zariski topology). (b) G is irreducible (as an algebraic variety): is not the union of two other varieties. (c) Ik (G) is a prime ideal. (2) The image of an algebraic group under a rational homomorphism is algebraic. (3) A group generated by connected algebraic subgroups is algebraic and connected (e.g., (a)–(g) are connected). It is defined over the perfect field k0 if each of the subgroups is. If G is an algebraic group, the radical of G (rad G) is the maximal connected solvable normal subgroup. G is semisimple if (1) rad G = {1} and (2) G is connected. Example.

has radical

⎧⎡ ⎪ ⎪ ⎪ ⎪⎢ ⎪ ⎪ ⎨⎢ ⎢ ⎢ ⎢ ⎪ ⎢ ⎪ ⎪ ⎣ ⎪ ⎪ ⎪ ⎩

1 ∗ − − − ∗ 0 | | A | 0 ⎧⎡ 1 ∗ − − ⎪ ⎪ ⎪ ⎪ ⎢ 1 ⎪ ⎪ ⎨⎢ ⎢ 1 ⎢ ⎢ 1 ⎪ ⎢ ⎪ ⎪ ⎣ ⎪ ⎪ ⎪ ⎩ 0

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬



⎥ ⎥ ⎥ ⎥ : A ∈ SLn−1 ⎥ ⎪ ⎥ ⎪ ⎪ ⎦ ⎪ ⎪ ⎪ ⎭ − ∗ 0 1 1

⎤⎫ ⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎥⎪ ⎥⎬ ⎥ ⎥⎪ ⎥⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎭

For the remainder of this section we assume that k is algebraically closed, k0 is the prime field, G is a Chevalley group based on k and M is the lattice. (Since a change of basis in M is given by polynomials with integral coefficients we may speak of a basis over M .) Theorem 6. With the preceding notations: (a) G is a semisimple algebraic group relative to M . (b) (c) (d) (e)

B is a maximal connected solvable subgroup (Borel subgroup). H is a maximal connected diagonalizable group (maximal torus). N is the normalizer of H and N/H ∼ = W. G, B, H, and N are defined over k0 relative to M .

Remark. B and H are determined up to conjugations by the abstract group G: (a) B is maximal solvable and has no subgroups of finite index. (b) H is maximal nilpotent and every subgroup of finite index is of finite index in its own normalizer. 14 See

also Borel [6, Chapter 1, §§ 1,2]. ()

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5. CHEVALLEY GROUPS AND ALGEBRAIC GROUPS

39

Proof of Theorem 6. (a) Map Ga → Xα by xα : t → xα (t). This is a rational homomorphism. So since Ga is a connected algebraic group so is Xα . Hence G is algebraic and connected. Let R = rad G. Since R is solvable and normal it is finite by the corollary to Theorem 5. Since R is also connected R = 1, and hence G is semisimple. (b) and (c). H is the image of Glm under (t1 , . . . , tl ) → li=1 hi (ti ) and hence is algebraic and connected; so B = U H is connected, algebraic and solvable. Let G1  B. Then G1 ⊇ Bwα B (some simple root α), so G1 ⊇ Xα , X−α , and hence by Corollary 6 of Theorem 4 G1 is not solvable and hence (b) holds. H is a maximal connected diagonalizable subgroup of B (for any larger subgroup must intersect U nontrivially). Hence H is a maximal connected diagonalizable subgroup of G (by a theorem in Chevalley’s S´eminaire); so (c) holds. (d) is clear. (e) It suffices by (3) to prove Lemma 34.  Lemma 34. Let Xα = {xα (t) : t ∈ k} and α = {hα (t) : t ∈ k∗ }. Then: (a) Xα is defined over k0 and xα : Ga → Xα is an isomorphism over k0 . (b) α is defined over k0 and hα : Gm → α is a homomorphism over k0 . Proof. Let {vi } be a basis vi so that of M formed of weight vectors. Choose cij vj , and choose vj so that cij = 0.15 If vi is of Xα vi = 0, then write Xα vi = weight μ, then vj is of weight μ + α. Since xα (t) = 1 + tXα + t2 Xα2 /2 + · · · it follows that if aij is the (i, j) matrix coordinate (i = j) function then aij (xα (t)) = cij t. All other coefficients of xα (t) are polynomials over k0 in t, hence also in aij . This set of polynomial relations defines Xα as an algebraic group over k0 . Now c1ij aij : xα (t) → t is an inverse of xα , so the map xα is an isomorphism over k0 . The proof of (b) is left as an exercise.  We can recover the lattices L0 and L from the group G as follows. Let μ ∈ L. μ(H ) Define μ

: H → Gm by μ

( hi (ti )) = ti i . This is a character defined over k0 .

the character group of H. The Xα ’s are determined by { μ} generates a lattice L, H as the unique minimal unipotent subgroups normalized by H. If h = hi (ti ) α(H ) α(h)t) where α

(h) = ti i . α

is called the global root. then hxα (t)h−1 = xα ( 



. Then L0 ⊂ L. Define L0 = the lattice generated by all α

such that L0 → L 0 , μ → μ Exercise. There exists a W -isomorphism: L → L

,

is given by the action of N/H on the character and α → α

. (The action of W on L group). We summarize our results in: Existence Theorem. Given a root system Σ, a lattice L with L0 ⊂ L ⊂ L1 (where L0 and L1 are the root and weight lattices, respectively), and an algebraically closed field k, then there exists a semisimple algebraic group G defined over k such that L0 and L are realized as the lattices of global roots and characters, respectively, relative to a maximal torus. Furthermore, G, Xα , . . . can be taken to be defined over the prime field. 15 In fact, we need that c ij = 0 in the field k. The possibility of such choice of cij follows from the fact that Xα (V k ) = 0 (see footnote on page 17). () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

40

5. CHEVALLEY GROUPS AND ALGEBRAIC GROUPS

The classification theorem, that up to k-isomorphism every semisimple algebraic group over k has been obtained above, is much more difficult. (See S´eminair Chevalley [51]). We recall that HZ = H ∩ LZ = {H ∈ H : μ(H) ∈ Z for all μ ∈ L}, where LZ is the stabilizer in L of M , the lattice used to construct our Chevalley group in the first place. Lemma 35. Let k be algebraically closed, G a Chevalley group over k, H1 , . . . , Hl  a basis for HZ . Define hi by hi (t)v = tμ(Hi ) v for all v ∈ Vμ . Then the map ϕ : Glm → H given by (t1 , . . . , tl ) → lj=1 hj (tj ) is an isomorphism over k0 of algebraic groups.    Proof. Write Hi = nij Hj , nij ∈ Z. Given tj we can find {ti } such n that tj = i ti ij (for det(nij ) = 0 and k∗ is divisible). Then j hj (tj ) acts on μ(H  ) μ(H ) Vμ as multiplication by j tj j = i ti i , i.e., as hi (ti ). This shows that l ϕ maps Gm onto H. Clearly  ϕ is a rational mapping defined over  k0 . Let {μi } be the basis of L dual to Hj (i.e., μi (Hj ) = δij ). Write μi = μ∈L nμ μ. Then μ(Hj ) !nμ = ti so ϕ−1 exists and is defined over k0 .  μ j tj Theorem 7. Let k be an algebraically closed field and k0 the prime subfield. Let G be a Chevalley group parametrized by k and viewed as an algebraic group defined over k0 as above. Then: (a) U − HU is an open subvariety of G defined over k0 . (b) If n is the number of positive roots, then the map ϕ : kn × (k∗ )l × kn → U − HU defined by    xα (tα ) hi (ti ) xα (tα ) ϕ ((tα )α0 ) = α0

is an isomorphism of varieties over k0 .

" Proof. (a) We consider the natural action of G on n L, the nth exterior of products of power of L,16 relative to a basis " {Y1 , Y2 , . . . , Yr } over k0 made up aij (x)Yj and Hi ’s and Xα ’s such that Y1 = α>0 Xα . For x ∈ G we set xYi = then d = a11 , a function on G over k0 . We claim that x ∈ U − HU = U − B if and Since B fixes Y1 up to a nonzero multiple and only if d(x) = 0. Assume x ∈ U − B. if u ∈ U − then uXα ∈ Xα + H + ht(β)0 xβ (tβ ). Choosing an ordering of the positive roots consistent with addition, we see at once that fi(α),j(α) (x) = n(α)tα + an integral polynomial of the earlier t’s and that fij (x) is an integral polynomial in the t’s for all i, j. Thus ψ3 is an isomorphism over k0 and similarly for ψ1 . To prove θ is an isomorphism we order the vi so that U − , H, U consist respectively of subdiagonal unipotent, diagonal, superdiagonal unipotent matrices (see Lemma 18, Corollary 3), and then we may assume that they consist of all of the invertible matrices of these types. Let x = u− hu be in U − HU and let the subdiagonal entries of u− , the diagaonal entries of h, the superdiagonal entries of u be labeled tij with i > j, i = j, i < j respectively. We order the indices so that ij precedes kl in case i ≤ k, j ≤ l and ij = kl. Then in the three cases above fij (x) = tij tjj , resp. tij , resp. tii tij , increased by an integral polynomial in t’s preceding tij . We may now inductively solve for the t’s as rational forms over Z in the f ’s, the division of the forms representing the tjj ’s being justified by the fact that they are nonzero on U − HU . Thus θ is an isomorphism over k0 and (b) follows.  Example. In SLn , U − HU consists of all (aij ) such that the minors ï ò a11 a12 [a11 ] , , ... a21 a22 are all nonsingular. We can now easily prove the following important fact (but will refer the reader to Chevalley [17] instead). Let G be a Chevalley group over C, viewed as above as an algebraic matrix group over Q, the prime field, and I the corresponding ideal over Z (consisting of all polynomials over Z which vanish on G). Then the set of zeros of I in any algebraically closed field k is just the Chevalley group over k of the same type (same root system and same weight lattice) as G. Thus we have a functorial definition in terms of equations of all of the semisimple algebraic groups of any given type. Corollary 1. Let k, k0 , G, V be as above. Let G be a Chevalley group constructed using V  instead of V but with the same L. Assume that LV ⊇ LV  . Then the homomorphism ϕ : G → G taking xα (t) → xα (t) for all α and t is a homomorphism of algebraic groups over k0 . Proof. Consider first ϕ|U − HU . By Theorem 7 we need only show that ϕ|H μ (H ) is rational over k0 . The nonzero coordinates of hi (ti ) are ti i (μ ∈ LV  ). μ(Hi ) The nonzero coordinates of hi (ti ) are ti (μ ∈ LV ). Each of the former is a monomial in the latter (because LV  ⊆ LV ), and hence is rational over k0 .  Now for w ∈ W , ωw (resp. ωw ) can be chosen with coefficients in k0 (for wα (1) = −1 − U B, xα (1)x−α (1)xα (1)), so that ϕ|ωw−1 U − B is rational over k0 . Since Bωw B ⊆ ωw 17  we conclude that ϕ is rational over k0 . Corollary 2. The homomorphism ϕα : SL2 → Xα , X−α  (of Corollary 6 to Theorem 4  ) is a homomorphism of algebraic groups over k0 . 17 Here we use the fact that a map that is rational in a neighborhood of every point is rational everywhere. The proof of this statement is given in Lemma 69 in Chapter 12. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

42

5. CHEVALLEY GROUPS AND ALGEBRAIC GROUPS



Proof. This is a special case of Corollary 1.

Corollary 3. Assume L, V and M are fixed, that V is universal, k ⊂ K are fields and Gk , GK are the corresponding Chevalley groups. Then Gk = GK ∩GLM,k . Proof. Clearly Gk ⊆ GK ∩ GM,k . Suppose x ∈ GK ∩ GLM,k . Then x = uhωw v (see Theorem 4 ) with ωw defined over the prime field. We must show that −1 −1 xωw ∈ Gk , i.e., uhu− ∈ Gk where u− = ωw vωw . Write    xα (tα ) hi (ti ) xα (tα ) uhu− = α>0

α0 × (ti ) × (tα )α 0. By (B) Xβ is normalized by Xα and X−α and hence by wα (t). Thus wα (t)xβ (u)wα (−t) ∈ XS .19 By Lemma 36 we need only prove that relation (a) holds in G. But in G (a) follows from the relations (R). Now assume α = β and rank Σ ≥ 2. In this case we use the fact (see the corollary to Lemma 33) (∗) There exist roots δ and η and a positive integer j such that α = δ + jη and     xmδ+nη (cm,n tm un ) and c1j = 0. xδ (t), xη (u) = m,n>0

Set T = {mwα δ + nwα η : m, n positive integers}. Transforming both sides of the above equation by wα (t) and applying the case of (a) already proved we see that the transform of every term except xδ+jη (c1j tuj ) is an element of XT . Hence wα (t)xδ+jη (c1j tuj )wα (−t) ∈ XT , so by the earlier argument, with T in place of S, (a) holds. If α = β and rank Σ = 1 then (a) holds by (B ). Since wα (t)−1 = wα (−t), the case α = −β follows from the case α = β.  (b)–(f) follow from (a) and the definitions of wα (t) and hα (t). Part (a) of Theorem 8 follows from parts (a)–(d) of Lemma 37. Lemma 38. Let α be the group generated by all hα (t), i = αi , and H  the group generated by all α . Then:20 (a) Each α is normal in H  . (b) H  = li=1 i . Proof. (a) follows from Lemma 37(f). (b) Let β be any root, and write β = wαi with αi simple and w ∈ W . Let w = wα · · · be a minimal expression for w as a product of simple reflections. Set γ = wα β. Then hβ (t) = wα (1) hγ (c(−1)− β,α t) hγ (c(−1)− β,α )−1 wα (−1) by Lemma 37(c), and hence by Lemma 37(e) hβ (t) = hγ (c(−1)− β,α t) hγ (c(−1)− β,α )−1 wα (t− β,α ) wα (−1) ∈ γ · α 19 Since 20 Here

 (t)−1 = w  (−t). () wα α we assume relations (A), (B), and (B ), but not necessarily (C). ()

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6. GENERATORS AND RELATIONS

45



By induction on the length of w, (b) follows.

Proof of Theorem 8(b). Let G be the group generated by {xα (t) : α ∈ Σ, t ∈ k} subject to the relations (A), (B) if rank Σ > 1 or (B ) if rank Σ = 1, and (C). Let wα (t), hα (t), . . . be defined as usual in terms of the xα (t). We wish to prove that of the proposition π  : G → G is an isomorphism. Let x ∈ ker π  . By Corollary 1 in Chapter 3 (see page 28), x ∈ H  . By Lemma 38 and (C) x = hi (ti ) (ti ∈ k∗ ). Applying π  we obtain 1 = hi (ti ). Since G is universal each ti = 1, so x = 1.  Remarks. (a) In (A) and (B) it is sufficient to use as generators xα (t) where α is a linear combination of 2 simple roots and the relations (A) and (B) which can be written in terms of such elements. (b) It is sufficient to assume (C) for one root in each orbit under W . So for the simple roots, is enough. Exercise. If Σ is indecomposable, prove that it is sufficient to assume (C) for any long root α. We will now show that if k is an algebraic extension of a finite field then (A) and (B) imply (C). Lemma 39. Let α be a root and G as above. In G set f (t, u) = hα (t) hα (u) hα (tu)−1 . Then: (a) f (t, u2 v) = f (t, u2 )f (t, v). (b) If t, u generate a cyclic subgroup of k∗ , then f (t, u) = f (u, t). (c) If f (t, u) = f (u, t), then f (t, u2 ) = 1. (d) If t, u = 0 and t + u = 1, then f (t, u) = 1. Proof. (a) Since f (t, u) ∈ ker π, f (t, u) ∈ Z(G ). Set hα (t) = h(t). Then f (t, v) = h(u)f (t, v) h(u)−1 = h(u) h(t) h(v) h(tv)−1 h(u)−1 = h(tu2 ) h(u2 )−1 h(vu2 ) h(u2 )−1 h(u2 ) h(tu2 v)−1 2 −1

2

= h(tu ) h(u ) 2 −1

= f (t, u )

2

2

h(vu ) h(tu v)

(by Lemma 37(f))

−1

f (t, u2 v)

which proves (a). (b) Let t = v m , u = v n with m, n ∈ Z+ . Then h(t) = h(v)m c, h(u) = h(v)n d with c, d ∈ Z(G ), since G is a central extension of G. Thus h(t), h(u) commute and f (t, u) = f (u, t). (by Lemma 37(f)), so that (c) h(t) = h(u) h(t) h(u)−1 = h(tu2 ) h(u2 )−1 f (t, u2 ) = 1. (d) Abbreviate xα , x−α , wα , hα to x, y, w, h, respectively. We have: (1) w(t) x(u) w(−t) = y(−t2 u) (2) w(t) y(u) w(−t) = x(−t2 u) (by (1)) Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

46

6. GENERATORS AND RELATIONS

(3) w(t) = x(t) y(−t−1 ) x(t) (3 ) w(t) = y(−t−1 ) x(t) y(−t−1 ) Then

(by (1), (2), (3))

h(tu) h(u)−1 = w(tu) w(−u)

(by definition of h)

= x(t) x(−t) w(tu) w(−u) = x(t) w(tu) y(t−1 u−2 ) w(−u) = x(t) y(−t

−1 −1

u

) x(tu) y(−t

(by (1))

−1 −1

u

) y(t

−1 −2

u

) w(−u)

(by (3 ))

= x(t) y(−t−1 u−1 ) x(tu) y(u−2 ) w(−u) by (A) for y(−t−1 u−1 ) y(t−1 u−2 ) = y(t−1 u−2 (1 − u)) = y(u−2 ), h(tu) h(u)−1 = x(t) y(−t−1 u−1 ) w(−u) y(−tu−1 ) x(−1)

(by (1) and (2))

= x(t) y(−t−1 u−1 ) y(u−1 ) x(−u) y(u−1 ) y(−tu−1 ) x(−1)

(by (3 ))

= x(t) y(−t−1 u−1 (1 − t)) x(t − 1) y(u−1 (1 − t)) x(−1) = x(t) y(−t−1 ) x(t) x(−1) y(1) x(−1) = w(t) w(−1) = h(t) proving (d).



Lemma 40. In a field k of finite odd order there exist elements t, u such that t and u are not squares and t + u = 1. Proof. If |k| = q there are (q + 1)/2 squares in k. Since ((q + 1)/2)  q the squares do not form an additive group, so we can find a, b, c so that a + b = c where a and b are squares and c is not. Then take t = a/c, u = b/c.  Theorem 9. Assume that Σ is indecomposable and that k is an algebraic extension of a finite field. Then the relations (A) and (B) (or (B  ) if rank Σ = 1) suffice to define the corresponding universal Chevalley group, i.e., they imply the relations (C). Proof. Let t, u ∈ k∗ . We must show f (t, u) = 1 where f is as in Lemma 39. By Lemma 39 (b and c) if either t or u is a square f (t, u) = 1. Assume that both are not squares. By Lemma 40 (applied to the finite field generated by t and u) t = r 2 t1 , u = s2 u1 with r, s ∈ k∗ , t1 + u1 = 1, t1 and u1 not squares. Then f (t, u) = f (t, s2 u1 ) = f (t, u1 ) = f (r 2 t1 , u1 ) = f (t1 , u1 ) = 1 by Lemma 39 (a and d).  Example. If n ≥ 3 and k is a finite field, the symbols xij (t) (1 ≤ i, j ≤ n, i = j, t ∈ k) subject to the relations: (A) xij (t)xij (u) = xij (t + u), (B) (xij (t), xjk (u)) = xik (tu) if i, j, k are distinct, (xij (t), xkl (u)) = 1 if j = k, i = l, define the group SLn (k). (Recall: xij (t), e.g., equals 1 + tEij .)

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https://doi.org/10.1090//ulect/066/08

CHAPTER 7

Central Extensions Our object is to prove that if π, G and G are as in Chapter 6, then (π, G ) is a universal central extension of G in a sense to be defined. The reader is referred to the lecturer’s paper [59] in Colloque sur la th´eorie des groupes alg´ebriques, Bruxelles, 1962 and for generalities to Schur’s papers [48, 49, 50]. Definition. A central extension of a group G is a couple (π, G ) where G is a group and π is a homomorphism of G onto G, and ker π ⊆ Z(G ). Examples. (a) π, G , G as in Chapter 6. (b) π : G → G the natural homomorphism of one Chevalley group onto another constructed from a smaller weight lattice. E.g., π : SLn → PSLn , π : Spn → PSpn , and π : Spinn → SOn . (c) π : G → G a topological covering of a connected topological group, i.e., π is a local isomorphism, carrying a neighborhood of 1 in G isomorphically onto one of G. We note that π is central since its kernel being a discrete normal subgroup of a connected group, is necessarily central. To see this, let N be a discrete normal subgroup of a connected group G and let n ∈ N . Since the map G → N given by g → gng −1 , g ∈ G, has a discrete and connected image, gng −1 = n for all g ∈ G. Definition. A central extension (π, E) of a group G is universal if for any central extension (π  , E  ) of G there exists a unique homomorphism21 ϕ : E → E  such that π  ϕ = π, i.e., the following diagram is commutative: / E EA AA AA  A π AA π  G ϕ

We abbreviate universal central extension by u.c.e. We develop this property in a sequence of statements. (i) If a u.c.e. exists, it is unique up to isomorphism. Proof. If (π, E) and (π  , E  ) are u.c.e. of G, let ϕ : E → E  and ϕ : E  → E be such that π  ϕ = π and πϕ = π  . Now ϕ ϕ : E → E and π(ϕ ϕ) = π. Hence ϕ ϕ is the identity on E by the uniqueness of ϕ in the definition of a u.c.e. Similarly,  ϕϕ is the identity on E  . (ii) If (π, E) is a u.c.e. of G, then E = DE and hence G = DG, where DH denotes the derived group for any group H. 21 If (π , E ) and (π , E ) are central extensions of a group G then a homomorphism ϕ : 1 1 2 2 E1 → E2 such that π1 = π2 ϕ will be called a covering. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 47

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7. CENTRAL EXTENSIONS

Proof. Consider the central extension (π  , E  ) where E  = E × E/DE and π (a, b) = π(a), a ∈ E, b ∈ E/DE. Now if ϕ1 (a) = (a, 1) and ϕ2 (a) = (a, a + DE), then π  ϕi = π, i = 1, 2, and hence ϕ1 = ϕ2 . Thus E/DE = 1 and E = DE.  

(iii) If G = DG and (π, E) is a central extension of G, then E = C · DE, where C is a central subgroup of E on which π is trivial. Moreover, DE = D2 E. Proof. We have πDE = D(πE) = DG = G. Hence, E = CDE where C = ker π. Also, DE = D(CDE) = D2 E.  If (π, E) in (iii) covers all central extensions of G, then (π, DE) is universal, and E = A · DE (direct product) for some subgroup A of ker π. (iv) Theorem. If G = DG, then G possesses a u.c.e. Proof. For each x ∈ G we introduce a symbol e(x). Let F be the group generated by {e(x) : x ∈ G} subject to the condition that e(x) e(y) e(xy)−1 commutes with e(z) for all x, y, z ∈ G. If π : e(x) → x, then by using induction on the length of an expression in F , we see that π extends to a central homomorphism of F onto G. (a) (π, F ) covers all central extensions of G. To see this, let (π  , E  ) be any central extension of G. Choose e (x) ∈ E  such that π  e (x) = x. Since π  is central, the e (x)’s satisfy the condition on the e(x)’s. Hence, there is a homomorphism ϕ : F → E  such that ϕe(x) = e (x), and thus π  ϕ = π. (b) If E = DF and π also denotes the restriction of π to E, then (π, E) covers all central extensions uniquely. By (iii), we have that (π, E) covers all central extensions. If (π  , E  ) is a central extension of G and if π  ϕ = π = π  ϕ , then ϕ(x)ϕ (x)−1 ∈ Z(E). Thus, ψ : x → ϕ(x)ϕ (x)−1 is a homomorphism of E into an  abelian group. Since E = DE, ψ is trivial and ϕ = ϕ . Remark. Part (a) shows that if G is any group then there is a central extension covering all others. (iv ) If (π, E) is a central extension of G which covers all others and if E = DE, then (π, E) is a u.c.e. (v) If π : E → F and ψ : F → G are central extensions, then so is ψπ : E → G, provided E = DE. Proof. If a ∈ ker ψπ, let ϕa be the map ϕa : x → (a, x) = axa−1 x−1 , x ∈ E. Now ϕa (x) ∈ Z(E), since π(a, x) = (πa, πx) = 1 because πa ∈ Z(F ). Now ϕa is a  homomorphism,22 so ϕa is trivial, and a ∈ Z(E). (vi) Exercise. In (v), (π, E) is a u.c.e. of F if and only if (ψπ, E) is a u.c.e. of G. Definition. A group G is said to be centrally closed if (id, G) is a u.c.e. of G. (vii) Corollary. If (π, E) is a u.c.e. of G, then E is centrally closed. (viii) If E is centrally closed, then every central extension ψ : F → E of E splits, i.e., there exists a homomorphism ϕ : E → F such that ψϕ = id. 22 Indeed,

(axa−1 x−1 )(aya−1 y −1 ) = axa−1 aya−1 y −1 x−1 = a(xy)a−1 (xy)−1 . ()

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7. CENTRAL EXTENSIONS

49

(ix) (π, E) is a u.c.e. of G if and only if every diagram of the form E _ _ _/ E  π

π

 G

ρ

 / G

can be uniquely completed, where (π  , E  ) is a central extension of G and ρ is a homomorphism. Proof. One direction is immediate by taking G = G and ρ = id. Conversely, suppose the diagram is given and (π, E) is a u.c.e. Let H be the subgroup of G×E  , H = {(x, e ) : ρx = π  e }. If ψ : H → G is given by ψ(x, e ) = x, then ψ is central. Since (π, E) is a u.c.e. of G, there is a unique homomorphism θ : E → H such that ψθ = π. Now the homomorphism ϕ : E → E  , ϕ = ψ  θ, where ψ  : H → E  is given by ψ  (x, e ) = e , satisfies ρπ = ρψθ = π  ψ  θ = π  ϕ. If ρπ = π  ϕ , then let θ  : E → H be given by θ  (e) = (π(e), ϕ (e)). Since ψθ  = π, we have θ  = θ and ϕ = ψ  θ  = ψ  θ = ϕ, proving the uniqueness of ϕ.  Definition. A linear (respectively projective) representation of a group G is a homomorphism of G into some GL(V ) (respectively PGL(V )). Since GL(V ) is a central extension of PGL(V ) we have the following result: (x) Corollary. If (π, E) is a u.c.e. of G, then every projective representation of G can be lifted uniquely to a linear representation of E. (xi) Topological situation: If G is a topological group one can replace the condition G = DG by G is connected, the condition (π, E) is a u.c.e. by (π, E) is a universal covering group in the topological sense, and the condition that G is centrally closed by G is simply connected in the above discussion and obtain similar results. Definition. If (π, E) is a u.c.e. of the group G, then we call ker π the Schur multiplier of G. If we write ker π = M (G) to indicate the dependence on G, then a homomorphism ϕ : G → G leads to a corresponding one M (ϕ) : M (G) → M (G ) by (ix). Thus M is a functor from the category of groups G such that G = DG to the category of abelian groups with the following property: If ϕ is onto, then so is M (ϕ). Remark. Schur used different definitions (and terminology) since he considered only finite groups but did not require that G = DG. If G = DG our definitions are equivalent to his. One of Schur’s results, which we shall not use, is that if G is finite then so is M (G). Theorem 10. Let Σ be an indecomposable root system and k a field such that |k| > 4 and, if rank Σ = 1, then |k| = 9. If G is the corresponding universal Chevalley group (abstractly defined by the relations (A), (B), (B  ), (C) of Chapter 6), if G is the group defined by the relations (A), (B), (B  ) (we use (B  ) only if rank Σ = 1), and if π is the natural homomorphism from G to G, then (π, G ) is a u.c.e. of G. Remark. There are exceptions to the conclusion if the restrictions on k are omitted. E.g., SL2 (4) and SL2 (9) are such. Indeed, SL2 (4) ∼ = PSL2 (5) and SL2 (5) Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

50

7. CENTRAL EXTENSIONS

is a nontrivial central extension of PSL2 (5). For SL2 (9) see Schur. It can be shown that the number of couples (Σ, k) for which the conclusion fails is finite. Proof of Theorem 10. Since |k| > 4, G = DG, G = DG , u.c.e.’s exist for both G and G . The conclusion becomes G is centrally closed, by the above remarks. We need only show that every central extension (ψ, E) of G splits, i.e., there exists θ : G → E so that ψθ = idG ; i.e., the relations defining G can be lifted to E. We may assume E = DE; but then (πψ, E) is a central extension of G by (v). We need only show (1) If (ψ, E) is a central extension of G, then the relations (A), (B), (B ) can be lifted to E. Let C = ker ψ, a central subgroup of E. We have: (2) A commutator (x, y) with x, y ∈ E depends only on the classes mod C to which x and y belong. Choose a ∈ k∗ so that c = a2 − 1 = 0. Then in G, (hα (a), xα (t)) = xα (ct) for all α ∈ Σ, t ∈ k. We define ϕxα (t) ∈ E (α ∈ Σ, t ∈ k) so that ψϕxα (t) = xα (t) and so that (3) (ϕhα (a), ϕxα (t)) = ϕxα (ct) and then ϕh’s (and later ϕw’s in terms of the ϕx’s by the same formulas which define the h’s and the w’s in terms of the x’s). Note that this choice is not circular because of (2). We shall show that the relations (A), (B), (B ) hold with ϕx’s in place of x’s. (4) (ϕh)(ϕxα (t))(ϕh)−1 = ϕ(hxα (t)h−1 ) for all h ∈ H, α ∈ Σ, t ∈ k. Set hxα (t)h−1 = xα (dt) with d ∈ k∗ . Conjugating (3) by ϕh, we get (ϕh  α (a), ϕxα (dt))  = (ϕh) (ϕxα (ct)) (ϕh)−1 , and the left side equals ϕxα (cdt) = ϕ hxα (ct)h−1 by (3). Similarly we have   (4 ) (ϕn) (ϕxα (t)) (ϕn)−1 = ϕ nxα (t)n−1 for all n ∈ N , α ∈ Σ, t ∈ k. (5) If α and β are roots, α + β = 0, and α + β is not a root, then ϕxα (t) and ϕxβ (u) commute for all t, u ∈ k.   Set (ϕxα (t)) (ϕxβ (u)) ϕxα (t)−1 = f (t, u) (ϕxβ (u)), t, u ∈ k, f (t, u) ∈ C. We must show f (t, u) = 1. Clearly, from the definitions we have (6) f is additive in both positions. (6a) Assume α = β. If (α, β) = 0, then f (t, u) = f (tv 2 , u) (v = 0) by Lemma 20(c) and by (4) with h = hα (v). If (α, β) > 0, then f (t, u) = f (tv d , u) where d = 4 − α, β β, α by Lemma 20(c) and by (4) with h = hα (v 2 )hβ (v − β,α ). In both cases, f (t(1 − v d ), u) = 1 by (6) for some d = 1, 2 or 3. Choose v so that v d − 1 = 0. Then we get f ≡ 1. (6b) Assume α = β and rank Σ > 1. If there is a root γ so that α, γ = 1, set h = hγ (v) in the preceding argument and obtain (∗) f (t, u) = f (tv, uv). Choose v so that v − v 2 = 0 and 1 − v + v 2 = 0. By (∗) and (6), f (t(v − v 2 ), u) = f (t, u/(v − v 2 )) = f (t, u/v)f (t, u/(1 − v)) = f (vt, u)f ((1 − v)t, u) = f (t, u) whence f (t(1 − v + v ), u) = 1 and f ≡ 1. If there is no such γ, then Σ is of type Cn and α is a long root. In this case, however, α = β + 2γ with β and γ roots. 2

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7. CENTRAL EXTENSIONS

Thus,

51

  (ϕxβ (t), ϕxγ (u)) = g (ϕxβ+γ (±tu)) ϕxβ+2γ (±tu2 )

with g ∈ C, by Lemma 33. Since ϕxα (v) commutes with all factors but the last by (5a), it also commutes with the last. (6c) Assume α = β and rank Σ = 1. At least we have f (t, u) = f (tv 2 , uv 2 ), t, u ∈ k, v ∈ k∗ , using h = hα (v) in the argument above. We may also assume that |k| is not a prime. If it were, then xα (t) and xα (u) would be powers of xα (1) and (5) would be immediate. Referring to the proof of (5b), we see it will suffice to be able to choose v so that v, 1 − v are squares and v − v 2 = 0, 1 − v + v 2 = 0. If k is finite of characteristic 2, this is possible since all elements of k are squares. 2  2  Otherwise, set v = 2w/(1 + w2 ) . Then 1 − v = (1 − w2 )/(1 + w2 ) , and we need only choose w so that 1 + w2 = 0, v − v 2 = 0, and 1 − v + v 2 = 0. Since at most 13 values of w are to be avoided and |k| ≥ 25 in the present case, this too is possible. This completes the proof of (5). (7) ϕ preserves the relations (A). The element    −1 x = ϕxα (tc−1 ) ϕxα (uc−1 ) ϕxα ((t + u)c−1 ) is in C, and hence the transform of x by hα (a) is x itself. However, by (3), (4), (5) −1 this transform is also x (ϕxα (t)) (ϕxα (u)) (ϕxα (t + u)) . (8) ϕ preserves the relations (B). We have

ä Ä   (ϕxα (t)) (ϕxβ (u)) ϕxα (t)−1 = f (t, u) ϕxiα+jβ (cij ti uj ) (ϕxβ (u))

where f (t, u) ∈ C. One proves f = 1 by induction on n, the number of roots of the form iα + jβ, i, j ∈ Z+ . If n = 0, this is just (5). If n > 0, the inductive hypothesis and (7) imply f satisfies (6), and then the argument in (5a) may be used. (9) ϕ preserves the relations (B ). This follows from (4 ). This completes the proof of Theorem 10.



Exercise. Assume L is the original Lie algebra with coefficients transferred by means of a Chevalley basis to a field k whose characteristic does not divide any Nα,β = 0. Assume Σ is indecomposable of rank > 1. Prove: (a) The relations [Xα , Xβ ] = Nα,β Xα+β , α + β = 0, form a defining set for L. Hint: define Hα = [Xα , X−α ] and show that the relations in Theorem 1 hold. (b) L = DL, the derived algebra of L. (c) Every central extension of L splits. Hint: parallel the proof of Theorem 10. Corollary 1. The relations (A), (B), (B  ) can be lifted to any central extension of G. Let G be the group generated by (A), (B), or (A), (B ), as in Theorem 10, for all the following corollaries. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Corollary 2. (a) G is centrally closed. Each of its central extensions splits. Its Schur multiplier is trivial. It yields the u.c.e. of all of all the Chevalley groups of the given type, and covers linearly all of the projective representations of these groups. (b) If k is finite or more generally an algebraic extension of a finite field, then (a) holds with G replaced by G. Proof. This follows from various generalities at the beginning of this section.  E.g., if k is finite, |k| > 4 and SL2 (9) is excluded, then SLn (k), Spn (k) and Spinn (k) all have trivial Schur multipliers, and the natural central extensions SLn → PSLn , Spn → PSpn , Spinn → SOn are all universal. Corollary 3. Assume G, G and π as above. If k∗ is infinite and divisible (u ∈ k∗ , n ∈ Z implies there exists v ∈ k∗ with v n = u), then the Schur multiplier of G, i.e., C = ker π, is also divisible. Proof. Elements of the form f (t, u) = hα (t)hα (u)hα (tu)−1 in G , α ∈ Σ generate C. We have f (t, vw2 ) = f (t, v)f (t, w2 ) by Lemma 39(a). By induction, we get f (t, w2n ) = f (t, w2 )n for arbitrary n. Since u ∈ k∗ we can find w ∈ k∗ such  that u = w2n , the proof is complete. Corollary 3a. If k∗ is infinite and divisible by a set of primes including 2, then C is also divisible by these primes. Corollary 3b. If k∗ is infinite and divisible, then any central extension of G by a kernel which is a reduced group (no divisible subgroups other than 1) is trivial, i.e., it splits. Proof. Let (ψ, E) be a central extension of G with ker ψ reduced. Since (π, G ) is a u.c.e. we have ϕ : G → E so that ψϕ = π. Since C = ker π is divisible, so is ϕC ⊆ ker ψ. Hence ϕC = 1 and ker ϕ ⊇ ker π. Thus, there is a homomorphism  θ : G → E so that θπ = ϕ. Therefore, ψθπ = π on G and ψθ = 1 on G. Corollary 3c. If k∗ is infinite and divisible, then any finite-dimensional projective representation of G can be lifted uniquely to a linear representation. Proof. Assume σ : G → PGL(V ). Since G = DG, we have σ : G → PSL(V ). Let ρ : SL(V ) → PSL(V ) be the natural projection. Since dim V is finite, we have ker ρ is finite and thus ker ρ is reduced. Consider the central extension (ψ, E) of G where E = {(x, y) : σx = ρy, x ∈ G, y ∈ SL(V )} ⊆ G × SL(V ) and ψ(x, y) = x, (x, y) ∈ E. Now ker ψ = 1 × ker ρ is reduced, so by Corollary 3b, we have θ : G → E with ψθ = 1 on G. If ψ  (x, y) = y, (x, y) ∈ E, then σ  = ψ  θ :  G → SL(V ) ⊂ GL(V ) with ρσ  = ρψ  θ = σψθ = σ. Example. Corollary 3c says, for example, that every finite-dimensional projective representation of SLn (C) can be lifted to a linear one. (The novelty is that the representation is not assumed to be continuous.) Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Theorem 11. If Σ is an indecomposable root system, if char k = p = 0, and if G = DG (i.e, we exclude |k| = 2, Σ of type A1 , B2 or G2 and |k| = 3, Σ of type A1 ), then (π, G  ), with G  as in Theorem 10, defined by (A), (B), or (A), (B  ), uniquely covers all central extensions of G for which the kernel has no p-torsion. Proof. By Theorem 10, we could assume |k| ≤ 4 or |k| = 9. However, the proof to follow does not use this assumption or Theorem 10. If (ψ, E) is a central extension of G such that C = ker ψ has no p-torsion, then we wish to show (A), (B) and (B ) can be lifted to E. (1) Assume C is divisible by p, i.e., every element is. Choose ϕxα (t) ∈ E so that ψϕxα (t) = xα (t) and (ϕxα (t))p = 1, α ∈ Σ, t ∈ k. We claim relations (A), (B) and (B ) hold on the ϕx’s. (a) If α, β are roots, α + β not a root, and α + β = 0, then ϕxα (t) and ϕxβ (u) commute, t, u ∈ k. We have (ϕxα (t)) (ϕxβ (u)) (ϕxα (t))−1 = f ϕxβ (u) with f ∈ C. Taking p-th powers, we get 1 = f p which implies f = 1 since C has no p-torsion. (b) The relations (A) hold. Taking p-th powers of (ϕxα (t)) (ϕxα (u)) = f ϕxα (t + u), f ∈ C, we get f = 1 as before. (c) Exercise. Relations (B) and (B ) also hold. (2) General case (of Theorem 11). (a) C can be imbedded in a divisible group C  that has no p-torsion. We have a homomorphism θ of a free abelian group F onto C. Now F ⊗Z Q is a divisible group, and we can identify F with F ⊗Z Z ⊂ F ⊗Z Q. Hence C = F/ ker θ ⊂ F ⊗Z Q/ ker θ = D, say, and D is a divisible group. Moreover, since C has no p-torsion, C ∩ Dp = 1 where Dp is the p-component of D. Thus, C projects faithfully into D/Dp = C  which is divisible and has no p-torsion. (b) Conclusion of proof. Form E  = E × C  , the direct product of E and  C with C amalgamated, and define ψ  : E  → G by ψ  (e, c ) = ψ(e), e ∈ E, c ∈ C  . Now (ψ  , E  ) satisfies the assumptions of (1) so the relations (A), (B), (B ) can be lifted to E  . However, by Lemma 32 , the lifted group is its own derived group and hence contained in E.  Corollary 1. Every projective representation of G in a field of characteristic p can be lifted to a linear one. Corollary 2. The Schur multiplier of G is a p-group. Proofs of Corollaries 1 and 2. These are easy exercises.



Since the kernel of the map π : G → G above turns out to be the Schur multiplier of G, its structure for k arbitrary is of some interest. The result is: Theorem 12 (Moore, Matsumoto). Assume Σ is an indecomposable root system and k is a field with |k| > 4. If G is the universal Chevalley group based on Σ and k, if G is the group defined by (A), (B), (B  ), and if π is the natural map from G to G with C = ker π, the Schur multiplier of G, then C is isomorphic to the abstract group A generated by the symbols f (t, u) = hα (t)hα (u)hα (tu)−1 in G , the group defined by (A), (B), or (A), (B  ), (t, u ∈ k∗ ) subject to the relations, which are not independent of each other: (a) f (t, u)f (tu, v) = f (t, uv)f (u, v), f (1, u) = f (u, 1) = 1 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(b) f (t, u)f (t, −u−1 ) = f (t, −1) (c) f (t, u) = f (u−1 , t) (d) f (t, u) = f (t, −tu) (e) f (t, u) = f (t, (1 − t)u) and in the case Σ is not of type Cn (n ≥ 1) (C1 = A1 ) the additional relation: (ab ) f is bimultiplicative. In this case relations (a)–(e) may be replaced by (ab ) and (c ) f is skew (d ) f (t, −t) = 1 (e ) f (t, 1 − t) = 1 The isomorphism is given by ϕ : f (t, u) → hα (t) hα (u) hα (tu)−1 , α a fixed long root. Remark. These relations are satisfied by the norm residue symbol in class field theory, which is a significant aspect of Moore’s work. Partial Proof. (1) If α is the group generated (in G ) by all hα (t), α a fixed long root, then C ⊆ α . We know that hα (t) hα (u) hα (tu)−1 , α ∈ Σ, t ∈ k∗ , form a generating set for C. Using the Weyl group we can narrow the situation to at most two roots α, β with α long, β short, and (α, β) > 0. Hence, β, α = 1 and (hα (t), hβ (u)) = hβ (tu) hβ (t)−1 hβ (u)−1 = hα (t)hα (u α,β ) hα (tu α,β )−1 by Lemma 37(f). This shows α will suffice. (2) ϕ is a mapping onto C. This follows from (1). (3) ϕ is a homomorphism. We must show that the relations hold if f (t, u) is replaced by hα (t) hα (u)hα (tu)−1 . The relations (a) are obvious. A special case (u = 1) of (e), has been shown in Lemma 39(d). The other relations (b), (c), (d) follow from the commutator relations connecting the h’s and the w’s. (3 ) Assume Σ is not of type Cn . In this case there is a root γ so that α, γ = 1. Thus f (t, v) = hγ (u)f (t, v) hγ (u)−1 = hα (tu) hα (u)−1 hα (uv) hα (tuv)−1 = f (t, u)−1 f (t, uv), or f (t, uv) = f (t, u)f (t, v). By relation (c), f (uv, t) = f (u, t)f (v, t). (4) ϕ is an isomorphism. This is done by constructing an explicit model for G .  Now let G be a connected topological group. A covering of G is a couple (π, E) such that E is a connected topological group and π is a homomorphism of E onto G which maps a neighborhood of 1 in E homeomorphically onto a neighborhood of 1 in G; i.e., which is a local isomorphism. A covering is universal if it covers all other covering groups. If (id, G) is a universal covering, we say that G is simply connected. Remarks. (1) A covering (π, E) of a connected group is necessarily central as was noted at the beginning of this section. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(2) If a universal covering exists, then it is unique and each of its coverings of other covering groups is unique. This follows from the fact that a connected group is generated by any neighborhood of 1. (3) If G is a Lie group, then a universal covering for G exists and simple connectedness is equivalent to the property that every continuous loop can be shrunk to a point. (See Chevalley [16] or Cohn [18].) Theorem 13. If G is a universal Chevalley group over C viewed as a Lie group, then G is simply connected. Before proving Theorem 13, we shall first state a lemma whose proof we leave as an exercise. and

Lemma 41. If t1 , t2 , . . . , tn are complex numbers such that |ti | < , i = 1, 2, . . . , n n i=1 ti = 0, there exits ti and tj such that |ti + tj | < .

Proof of Theorem 13. Let (π, E) be a covering of G. Locally π is invertible so we may set ϕ = π −1 on some neighborhood of 1 in G. We shall show that ϕ can be extended to a homomorphism of G onto E, i.e., (id, G) covers (π, E). It suffices to show that ϕ can be extended to all of G so that the relations (A), (B), (B ), (C) hold on the ϕx’s. Consider the relations (A) (ϕxα (t)) (ϕxα (u)) = ϕxα (t + u), α ∈ Σ. Since ϕ is locally an isomorphism, there is an  > 0 such that (A) holds for |t| < ,  |u| < . If t ∈ k, t = ti , |ti | < , then set ϕxα (t) = ϕxα (ti ). Using induction and Lemma 41, we see that ϕxα (t) is well-defined. Clearly, (A) then holds for all t, u ∈ k. Alternatively, we could note that Xα is topologically equivalent to C and hence simply connected. Thus, ϕ extends to a homomorphism of Xα into E and (A) holds. Clearly the extension of ϕ to Xα is unique. To obtain the relations (B), let α, β be roots α = ±β, let S be the set of roots of the form iα+jβ (i, j ∈ Z+ ), and let XS be the corresponding unipotent subgroup of G. Topologically, XS is equivalent to Cn for some n, and is hence simply connected. As before ϕ can be extended to a homomorphism of XS into E, and the relations (B) hold. This extension is consistent with those above, by the uniqueness of the latter. We now consider hα (t) = xα (t) x−α (−t−1 )xα (t). We have wα (−1) = xα (t − 1) x−α (1 − t−1 )xα (−1) xα (t − 1)wα (−1) where xy = y −1 xy. Hence, if t is near 1 in C, then ϕhα (t) is near 1 in E. Thus, ϕhα (t) is multiplicative near 1 and hence abelian everywhere (recall ψ is central). We then have   −1 −1 ϕhα (u) = (ϕhα (t)) (ϕhα (u)) (ϕhα (t)) = ϕhα (t2 u) ϕhα (t2 ) by Lemma 37(f). Since C has square roots, we have ϕhα is multiplicative, i.e., (C) holds.  Examples. SLn (C), Spn (C), and Spinn (C) are simply connected. These cases can also be proved by induction on n. (See Chevalley [16, Chapter II]) Remarks. (a) If C is replaced by R in the preceding discussion, then relations (A), (B) can be lifted exactly as before. Also ϕhα is still multiplicative if one of the Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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two arguments is positive. Further ker π is generated by (ϕhα (−1))2 , α a fixed long 4 8 root, and, if type Cn (n ≥ 1) is excluded, then (hα (−1)) = 1 or (wα (−1)) = 1. Prove all of this. (b) Moore has constructed a universal covering of G and has determined the fundamental group in case k is a p-adic field, using appropriate modifications of the definitions (here G is totally disconnected). (c) Let G be a Chevalley group over k, G the corresponding universal group, and π : G → G the natural homomorphism. If k is algebraically closed and if only appropriate coverings are allowed, then (π, G ) is a universal covering of G in the sense of algebraic groups. We close this chapter with a result in which the coefficients may come from any ring (associative with 1). The development is based in part on a letter from J. Milnor. Let R be the ring, and let GL(R) be the group of infinite matrices which are equal to the identity everywhere except for a finite invertible block in the upper left hand corner. Thus GLn (R) ⊂ GL(R), n = 1, 2, . . .. Let E(R) be the subgroup of GL(R) generated by the elementary matrices 1 + tEij (t ∈ R, i = j, i, j = 1, 2, . . .), where Eij is the usual matrix unit. For example, if R is a field, then E(R) = SL(R), a simple group whose double coset decomposition involves the infinite symmetric group. Indeed, if R is a Euclidean domain,23 then E(R) = SL(R). Lemma 42. (a) E(R) = DGL(R). (b) E(R) = DE(R). Proof. The relations (1 + tEik , 1 + Ekj ) = 1 + tEij shows (b) and hence also E(R) ⊆ DGL(R). If x, y ∈ GLn (R), then xyx−1 y −1 ∈ E(R) because in GL2n (R) we have ï ò ï ò òï òï xyx−1 y −1 0 x 0 0 y 0 (yx)−1 (1) = 0 1 0 x−1 0 yx 0 y −1 òï òï ò ï ò ï òï 1 −1 1 0 x 0 1 x 1 0 (2) = 0 1 1−x 1 0 x−1 0 1 1 − x−1 1 ï ò  2n n  1 x (3) = (1 + xij Eij ), if x = (xij ).  0 1 i=1 j=n+1

We call K1 (R) = GL(R)/E(R) the Whitehead group of G. This concept is used in topology. The case in which R = Z[G] is of particular interest. Example. If R is a Euclidean domain, then K1 (R) = R∗ , the group of units. (See Milnor [44].)24 By Lemma 42 and (iv), E(R) has a u.c.e. (π, U (R)). Set K2 (R) = ker π. This notation is partly motivated by the following exact sequence. 1 → K2 (R) → U (R) → GL(R) → K1 (R) → 1 23 I.e.,

a commutative ring where one can use the Euclid algorithm. The precise definition can be found in any text on commutative algebra. () See also Hungerford [32]. 24 This also true in the case where R is a commutative local ring, or R is a ring of polynomials in n variables over an Euclidean ring (see [3]), or R is a Dedekind ring (see [4]). () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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K2 is a functor from rings to abelian groups with the following property: if R → R is onto, then so is the associated map K2 (R) → K2 (R ). Remark. K2 is known to the lecturer in the following cases: (a) If R is a finite field (or an algebraic extension of a finite field), then K2 = 1. (b) If R is any field,25 see Theorem 12. (c) If R = Z, then |K2 | = 2. Here (a) follows from Theorem 9 and the next theorem, and a proof of (c) will be sketched after the remarks following the corollaries to the next theorem. Theorem 14. Let U (R) be the abstract group generated by the symbols xij (t) (t ∈ R, 1 = j, i, j = 1, 2, . . .) subject to the relations (A) xij (t) is additive in t. ® xij (tu), if k = l, i = j, (B) (xik (t), xlj (u)) = 1, if k = l, i = j. If π : U (R) → E(R) is the homomorphism given by xij (t) → 1+tEij , then (π, U (R)) is a u.c.e. for E(R). Proof. (a) π is central. If x ∈ ker π, choose n large enough so that x is a product of xij ’s with i, j < n. Let Pn be the subgroup of U (R) generated by the ’s (k = n, k = 1, 2, . . .). Now by (A) and (B), any element of Pn can be expressed xkn as k xkn (tk ). Since in E(R) this form is unique, π|Pn is an isomorphism. Also by (A) and (B), xij (t)Pn xij (t)−1 ⊆ Pn if i, j < n. Thus, xPn x−1 ⊆ Pn . If y ∈ Pn , then π(x, y) = 1, and since π|Pn is an isomorphism we have (x, y) = 1. In particular, x commutes with all xkn (t). Similarly, x commutes with all xnk (t) and hence with all xij (t) = (xin (t), xnj (1)). Thus x is in the center of U (R). (b) π is universal. From (B) it follows that U (R) = DU (R). Hence it suffices to show it covers all central extensions. Let (ψ, A) be a central extension of E(R) and let C be the center of A. We must show that we can lift the relations (A) and (B) to A. Fix i, j, i = j and choose p = i, j. Choose yij (t) ∈ ψ −1 xij (t) so that (∗) (yip (t), ypj (1)) = yij (t). We will prove that the y’s satisfy the equations (A) and (B). (b1) If i = j, k = l, then yik (t) and ylj (u) commute. Choose q = i, j, k, l and write ylj (u) = c(ylq (u), yqj (1)), c ∈ C. Since yik (t) commutes up to an element of C with ylq (u) and yqj (1), it commutes with ylj (u). Hence (b2) {yij (t)}, i, j fixed, is abelian. (b3) The relations (A) hold. The proof is exactly the same as that of statement (7) in the proof of Theorem 10. (b4) yij (t) in (∗) is independent of the choice of p. If q = p, i, j, set w = yqp (1)ypq (−1)yqp (1). Transforming (∗) by w and using (b1), we get (∗) with q in place of p. (b5) The relations (B) hold. We will use: R is a field that is not an algebraic extension of either the field Q or of a finite field, then K2 (R) = 1. This and other properties of K2 are proved in a recent paper by Bass and Tate, see [5]. () 25 If

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(∗∗) If a, b, c are elements of a group such that a commutes with c and such that (b, c) commutes with (a, b) and c, then (a, (b, c)) = ((a, b), c). Since a commutes with c, (a, (b, c)) = ((a, b), (b, c)c). The other conditions insure ((a, b), (b, c)c) = ((a, b), c). Now assume i, j, k are distinct. Choose q = i, j, k, so that (yik (t), ykj (u)) = (yik (t), (ykq (u), yqj (1))) = ((yik (t), ykq (u)) , yqj (1)) = (yiq (tu), yqj (1)) = yij (tu) by (∗), (∗∗), and (b4). This completes the proof of the theorem.



Let Un (R) denote the subgroup of U (R) generated by yij (t) with i, j ≤ n. Corollary 1. If n ≥ 5, then Un (R) is centrally closed. Corollary 2. If R is a finite field and n ≥ 5 then SLn (R) is centrally closed. Proof. This follows from Corollary 1 and the equations En (R) = SLn (R) =  Un (R). Remarks. (a) It follows that if R is a finite field and if SLn (R) is not centrally closed, then either |R| = 9, n = 2 or |R| ≤ 4 and n ≤ 4. The exact set of exceptions is: SL2 (4), SL2 (9), SL3 (2), SL3 (4), SL4 (2). Exercise. Prove this. (b) The argument above can be phrased in terms of roots, etc. As such, it carries over very easily to the case in which all roots have one length. The only other exception is D4 (2). (c) By a more complicated extension of the argument, it can also be shown that the universal Chevalley group of type Bn or Cn over a finite field (or an algebraic extension of a finite field) is centrally closed if n is large enough. Hence, only a finite number of universal Chevalley groups with Σ indecomposable and k finite fail to be centrally closed. Now we sketch a proof that K2 (Z) is a group of order 2. The notation U, Un , . . . above will be used. The proof depends on the following result: (1) For n ≥ 3, SLn (Z) is generated by symbols xij (i, j = 1, 2, . . . , n; i = j) subject to the relations ® xij , if k = l, i = j, (B) (xik , xlj ) = 1, if k = l, i = j. −1 2 , then h2ij = 1. (C) If wij = xij xji xij , hij = wij Identifying xij with the usual xij (1) and using xij (t) = xij (1)t , we see that the relations (B) here imply those of Theorem 14. Since the last relation may be written hij (−1)2 = hij (1) and ±1 are the only units of Z, we have SLn (Z) defined by the usual relations (A), (B), (C) of Chapter 6. Perhaps there are other rings, e.g., the p-adic integers, for which this result holds. For the proof of (1) see Magnus [43], which gives the reference to Nielsen, who proved the key case n = 3 (it takes some work to cast Nielsen’s result into the above form). The case n = 2, with −1 = x−1 (B) replaced by (B ) w12 x12 w12 21 , is simpler and is proved in an appendix to Kurosh, Theory of Groups [39]. Now let xij , wij , hij refer to elements of Un (Z). Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(2) If Cn is the kernel of πn : Un (Z) → SLn (Z) and n ≥ 3, then Cn is generated by h212 and (h212 )2 = 1. 2 h212 = 1 As usual, we only require 2 (C) when i, j = 1, 2, and h12 ∈ Cn . Setting −1 amounts to dividing by h12 , which thus equals Cn . The relation h23 h12 h23 = h−1 12 , which may be deduced from (B) as in the proof of Lemma 37, then yields h212 = h−2 12 . 2 (3) h12 = 1 if n ≥ 3. Assume not. There is a natural map Un (Z) → Un (R), xij → xij (1). This maps h212 onto h12 (−1)2 , which (see Remark (a) after the proof of Theorem 13) generates the kernel of Un (R) → SLn (R). Thus SLn (R) is centrally closed, hence simply connected. Since SLn (R) can be contracted to SOn (by the polar decomposition, which will be proved in the next section) which is not simply connected since Spinn → SOn is a nontrivial covering, we have a contradiction. It now follows from (2), (3) and Theorem 14 that |K2 (Z)| = 2. By Corollary 1 above the same conclusion holds with SL(Z) replaced by any SLn (Z) with n ≥ 5. Exercise. Let SAn be SLn × translations of the underlying space, kn with k again a field. I.e., SAn is the group of all (n + 1) × (n + 1) matrices of the form ï ò x y 0 1 where x ∈ SLn , y ∈ kn . SAn is generated by xij (t), t ∈ k, i = j, i = 1, 2, . . . , n, j = 1, 2, . . . , n + 1. Prove: (1) If the relation (C) hij (t) is multiplicative is added to the relations (A) and (B) of Theorem 14, a complete set of relations for SAn is obtained. (2) If k is finite, (C) may be omitted. (3) If n is large enough, the group defined by (A) and (B) is a u.c.e. for SAn . (4) Other analogues of results for SLn . We remark that SA2 (C) is the universal covering group of the inhomogeneous Lorentz group, hence is of interest in quantum mechanics.

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https://doi.org/10.1090//ulect/066/09

CHAPTER 8

Variants of the Bruhat Lemma Let G be a Chevalley group, k, B, . . . as usual. We recall (Theorems 4 and 4 ):  (a) G = w∈W BwB (disjoint union) (b) For each w ∈ W , BwB = BwUw , with uniqueness of expression on the right. Our purpose is to present some analogues of (b) with applications. For each simple root α we set Gα = Xα , X−α , a group of rank 1, Bα = B ∩Gα , and assume that the representative of wα in N/H, also denoted wα , is chosen in Gα . Theorem 15. For each simple root α let Yα be a system of representatives for Bα \(Gα − Bα ), or more generally for B\Bwα B. For each w ∈ W choose a minimal expression w = wα wβ · · · wδ as a product of reflections relative to simple roots α, β, . . .. Then BwB = BYα Yβ · · · Yδ with uniqueness of expression on the right. Proof. Since Gα − Bα = Bα wα Bα , the second case above is really more general than the first. We have BwB = Bwα Bwα wB

(by Lemma 25)

= Bwα BYβ · · · Yδ

(by induction)

= BYα Yβ · · · Yδ Now assume b yα yβ · · · yγ yδ =

b yα yβ

(by the choice of Yα ). · · · yγ yδ

with b, b ∈ B, etc. Then

byα · · · yγ = b yα · · · yγ yδ yδ−1 . We have yδ yδ−1 ∈ B or Bwδ B. The second case cannot occur since then the left side would be in Bwwδ B and the right side in BwB (by Lemma 25). From the definition of Yδ it follows that yδ = yδ , and then by induction that yγ = yγ , · · · , whence the uniqueness in Theorem 15.  Lemma 43. Let ϕα : SL2 → Gα be the canonical homomorphism (see Theorem 4  , Corollary 6). Then Yα satisfies the conditions of Theorem 15 in each of the following cases: (a) Yα = wα Xα . (b) k = C (resp. R) and Yα is the image under ϕα of the elements of SU2 (resp. SO2 ) (standard compact forms) of the form ò ï a −b a b with b > 0. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 61

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(c) If θ is a principal ideal domain (commutative with 1), θ ∗ is the group of units, k is the quotient field, and Yα is the image under ϕα of the elements of SL2 (θ) of the form ï ò a b c d with c running through a set of representatives for (θ − {0})/θ ∗ , and for each c, a running over a set of representatives for the residue classes of θ mod c. Proof. We have (a) by Theorem 4 applied to Gα . To verify (b) and (c) we may assume that Gα is SL2 and Bα the superdiagonal subgroup B2 since ker ϕα ⊆ B2 . Any element of SL2 (C) can be converted to one of SU2 by adding a multiple of the second row to the first and normalizing the lengths of the rows. Thus SL2 (C) = B2 (C).SU2 . Then B2 (C)\SL2 (C) ∼ (B2 (C) ∩ SU2 )\SU2 , whence (b). Now assume ï ò p q ∈ SL2 (k) r s with k as in (c). We choose a, c ∈ θ relatively prime and such that ra + sc = 0 (using unique factorization), and then b, d ∈ θ so that ad − bc = 1. Multiplying the preceding matrix on the right by ï ò a b c d we get an element of B2 (k). Thus SL2 (k) = B2 (k)SL2 (θ), and (c) follows.



Remarks. (a) The case (a) above is essentially Theorem 4 since wUw = wα Xα .wβ Xβ · · · wδ Xδ in the notation of Theorem 15, by Appendix II 25, or else by induction on the length of the expression. (b) In (c) above the choice can be made precise in the following cases: 1. θ = Z; choose a, c so that 0 ≤ a < c. 2. θ = F [X] (F a field); choose so that c is monic and deg a < deg c. 3. θ = Zp (p-adic integers); choose c a power of p and a an integer such that 0 ≤ a < c. In what follows we will give separate but parallel developments of the consequences of (b) and (c) above. In (b) we will treat the case k = C for definiteness, the case k = R being similar. Lemma 44. Let L and {Xα , Hα } be as in Theorem 1. (a) There exists an involutory semiautomorphism σ0 of L (relative to complex conjugation of C) such that σ0 Xα = −X−α and σ0 Hα = −Hα for every root α. (b) On L the form {X, Y } defined by (X, σ0 Y ) in terms of the Killing form is negative definite. Proof. This basic result is proved, e.g., in Jacobson [35, p. 147].



Theorem 16. Let G be a Chevalley group over C viewed as a Lie group over R. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

8. VARIANTS OF THE BRUHAT LEMMA

63

(a) There exists an analytic automorphism σ of G such that σxα (t) = x−α (−t) −1 and σhα (t) = hα (t ) for all α and t. (b) The group K = Gσ of fixed points of σ is a maximal compact subgroup of G and the decomposition G = BK holds (Iwasawa decomposition). Proof. Let σ1 be σ0 in Lemma 44 composed with complex conjugation, and ρ the representation of L used to define G. Applying Theorem 4 , Corollary 5, to the Chevalley groups (both equal to G) constructed from the representations ρ and ρ ◦ σ1 of L, we get an automorphism of G which aside from complex conjugation satisfies the equations of (a), hence composed with conjugation satisfies these equations. From Theorem 7 adapted to the present situation (see remark at the end of Chapter 5) it follows that σ is analytic, whence (a). We observe that if G is defined by the adjoint representation of L, then σ is effected by conjugation by the  semiautomorphism σ0 of Lemma 44. Lemma 45. Let K = Gσ , Kα = K ∩ Gα for each simple root α. (see Lemma 43(b)), hence Yα ⊂ Kα .

(a)

Kα = ϕα SU2

(b)

B σ = Hσ ¶ ©

(global weights) = h ∈ H : | μ(h)| = 1 for all μ

∈L © ¶ (see Lemma 28) = hi (ti ) : |ti | = 1 = maximal torus in K.

Proof. The kernel of ϕα : SL2 → Gα is contained in {±1}, and σ pulls back to the inverse transpose conjugate, say σ2 , on SL2 . Since the equation σ2 x = −x has no solutions we get (a). −1 Since σhα (t) = hα (t ), μ

(hα (t)) = tμ(Hα ) (here μ and μ

are the corresponding −1

weights on H and H), and the hα (t) generate H, we have μ

(σh) = μ

(h) for all h ∈ H, so that σh = h if and only if | μ(h)| = 1 for all weights μ

. If h = hi (ti ), μ(H ) then μ

(h) = ti i . Since there are l linearly independent weights μ, we see that n if | μ(h)| = 1 for all μ

, then |ti | = 1 for some n > 0, whence |ti | = 1, for all i. If G is universal, then Bσ is the product of the l circles {hi (·)}, hence is a torus; if not, we have to take the quotient by a finite group, thus still have a torus. Now

(h) (α ∈ Σ) are distinct and if h ∈ Hσ is general enough, so that the numbers α different from 1, then Gh , the centralizer of h in G, is H, by the uniqueness in Theorem 4 , so that Hσ is in fact a maximal abelian subgroup of Gσ , which proves the lemma.  Exercise. Check out the existence of h and the property of Gh = H above. Now we consider part (b) of Theorem 16. By Theorem 15 and Lemma 43(b) and 45(a) we have G = BK. By the same results (BwB)σ ⊆ Bσ Kα · · · Kδ , a comapct set since each factor is (the compactness of tori and SU2 is being used). Thus K = Gσ is compact. (This also follows easily from Lemma 44(b)). Let K1 be a compact subgroup of G, K1 ⊇ K. Assume x ∈ K1 . Write x = by with b ∈ B, y ∈ K, and then b = uh with u ∈ U , h ∈ H. Since K1 is compact, all eigenvalues μ

(hn ) (n = 0, ±1, ±2, · · · ) are bounded, whence h ∈ K by Lemma 45(b). Then all coefficients of all un are bounded so that u = 1. Thus x ∈ K, so that K is maximal compact. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Remark. It can be shown also that K is semisimple and that a complete set of semisimple compact Lie groups is got from the above construction. Corollary 1. Let G be of the same type as G with a weight lattice containing that of G, K  = Gσ , and π : G → G the natural projection. Then πK  = K. Proof. This follows from the fact proved in Lemma 45 that K is generated by the groups ϕα SU2 .  Examples. (a) If G = SLn (C), then K = SUn .  xi xn+1−i and (b) If G = SOn (C), then K fixes simultaneously the forms  xi xi , hence equals SOn (R) (compact form) after a change of coordinates. Prove this. n  (c) If G = Sp2n (C), then K fixes the forms i=1 (xi y2n+1−i − x2n+1−i yi ) and xi xi , and is isomorphic to SUn (H) (compact form, H = quaternions). For this see Chevalley [16, p. 22]. (d) We have isomorphisms and central extensions, ∼ SU2 (C) → SO3 (R) H∗ = SU1 (H) = SU2 (H) → SO5 (R) SU2 (C)2 → SO4 (R) SU4 (C) → SO6 (R) (compact forms). This follows from (a), (b), (c), Corollary 1 and the equivalences C1 = A1 = B1 , C2 = B2 , A21 = D2 , A3 = D3 . Corollary 2. The group K is connected. Proof. As already remarked, K is generated by the groups ϕα SU2 . Since SU2 is connected, so is K.  Corollary 3. If T denotes the maximal torus Hσ , then T \K is homeomorphic to B\G under the natural map. Proof. The map K → B\G, k → Bk, is continuous and constant on the fibers of T \K, hence leads to a continuous map of T \K into B\G which is 1–1 and onto since T = B ∩ K and G = BK. Since T \K is compact, the map is a homeomorphism.  Corollary 4. (a) G is contractible to K. (b) If G is universal, then K is simply connected. ¶ ©

. Then we have H = AT , so Proof. Let A = h ∈ H : μ

(h) > 0 for all μ

∈L that G = BK = U AK. On the right there is uniqueness of expression. Since K is compact it easily follows that the natural map U A × K → G is a homeomorphism. Since U A is contractible to a point, G is contractible to K. If also G is universal, then G is simply connected by Theorem 13; hence so is K.  Corollary 5. For w ∈ W set (BwB)σ = BwB ∩ K = Kw , and let α, β, . . . , δ be as in Theorem 15. Then K = ∪w Kw and Kw = T Yα . . . Yδ , with uniqueness of expression on the right. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

8. VARIANTS OF THE BRUHAT LEMMA

Proof. This follows from Theorem 15 and Lemma 43(b).

65



Remark. Observe that Kw is essentially a cell since each Yα is homeomorphic to C (consider the values of a in Lemma 43(b)). A true cellular decomposition is obtained by writing T as a union of cells. Perhaps this decomposition can be used to give an elementary treatment of the cohomology of K.  Corollary 6. B\G and T \K have as their Poincar´e polynomials w∈W t2N (w) . They have no torsion. Proof. We have B\BwB homeomorphic to wUw , a cell of real dimension 2N (w). Since each dimension is even, it follows that the cells represent independent elements of the homology group and that there is no torsion (essentially because the boundary operator lowers dimensions by exactly 1), whence Corollary  6. Alternately one may use the fact that each Yα is homeomorphic to C. Remark. The above series will be summed in the next section, where it arises in connection with the orders of the finite Chevalley groups. Corollary 7. For w ∈ W let w = wα · · · wδ be a minimal expression as before and let S denote the set of elements of W each of which is a product  of some subsequence of the expression for w. Then K w (topological closure) = w ∈S Kw . Proof. If Tα = T ∩ Kα , we have Kα = Tα Yα ∪ Tα by Lemma 45(a) and Tα Yα = Kα by the corresponding result in SU2 . Now BwB = B · Tα Yα · · · Tδ Yδ by Lemma 43(b). Hence Kw = T · Tα Yα · · · Tδ Yδ , so that K w ⊇ T Kα · · · Kδ , and we have equality since each factor on the right is compact, so that the right side is compact, hence closed. Since Kw Kα ⊆ Kw ∪ Kw wα if w ∈ W and α is simple, by Lemma 25, Corollary 7 follows.  Corollary 8. (a) T = K1 is in the closure of every Kw . (b) Kw is closed if and only if w = 1. Corollary 9. The set S of Corollary 7 depends only on w, not on the minimal expression chosen, hence may be written S(w). Proof. Because K w does not depend on the expression.



Lemma 46. Let w0 be the element of W which makes all positive roots negative. Then S(w0 ) = W . Proof. Assume w ∈ W , and let w = w1 · · · wm be a minimal expression as a product of simple reflections and similarly for w−1 w0 = wm+1 · · · wn . Then w0 = w1 · · · wm · · · wn is one for w0 since if N is the number of positive roots then m = N (w), n = N − N (w), and m + n = N = N (w0 ). Looking at the initial  segment of w0 we see that w ∈ S(w0 ). Corollary 10. If w0 is as above and w0 = wα wβ · · · wδ is a minimal expression, then (a) K = K w0 . (b) K = Kα Kβ · · · Kδ . Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Proof. (a) By Corollary 7 and Lemma 46. Tγ (γ simple), then (b) By (a) K = T Kα Kβ · · · Kδ . We may write T = absorb the Tγ ’s in appropriate Kγ ’s to get (b).  Exercise. If G is any Chevalley group and w0 , α, β, . . . are as above, show that G = BGα Gβ · · · Gδ . Remarks. (a) If C and SU2 are replaced by R and SO2 in accordance with Lemma 43(b), then everything goes through except for Corollary 4, Corollary 6, and the fact that T is no longer a torus. In this case each Kα is a circle since SO2 is. The corresponding angles in Corollary 10(b), which we have to restrict suitably to get uniqueness, may be called Euler angles in analogy with the classical case: G = SL3 (R)

K = SO3 (R)

Kα , Kβ = {rotations around the z-axis, x-axis} K = K α Kβ Kα . (b) If Kw is replaced by BwB = BKw in Corollary 7, the formula for BwB is obtained. (Prove this.) If C (or R) is replaced by any algebraically closed field and the Zariski topology is used, the same formula holds. So as not to interrupt the present development, we give the proof later, at the end of this section. Theorem 17 (Cartan). Again let G be a Chevalley ¶ © group over C or R, K = Gσ

. as above, and A = h ∈ H : μ

(h) > 0 for all μ

∈L (a) G = KAK (Cartan decomposition). (b) In (a) the A-component is determined uniquely up to conjugacy under the Weyl group. Proof. (a) Assume x ∈ G. By the decomposition H = AT and G = BK (Theorem 16), there exist elements in KxK ∩ U A. Given such an element y = ua, we write a = exp H (H ∈ HR , uniquely determined by a), then set |a| = |H|, the Killing norm in LR . This norm is invariant under W . We now choose y to maximize |a| (recall that K is compact). We must show that u = 1. This follows from: if u = 1, then |a| can be increased. We will reduce (∗) to the rank 1 case. Write u = β>0 uβ (uβ ∈ Xβ ). We may assume uα = 1 for some simple α: choose α of minimum height, say n, such that uα = 1, then if n > 1, choose β simple so that (α, β) > 0 and ht wβ α < n, then replace y by wβ (1)ywβ (1)−1 and proceed by induction on n. We write u = u uα with u ∈ XP −{α} (here P is the set of positive roots). Then we write a = exp H, choose c so that H  = H −cHα is orthogonal to Hα , set aα = exp cHα ∈ A∩Gα , a = exp H  ∈ A, a = aα a . Then a commutes with Gα elementwise and is orthogonal to aα relative to the bilinear form corresponding to the norm introduced above. By (∗) for groups of rank 1, there exist y, z ∈ Kα such that yuα aα z = aα ∈ A ∩ Gα and |aα | > |aα |. Then yuaz = yu uα aα a z = yu y −1 aα a . Since Gα normalizes XP −{α} 2 2 2 2 2 (since Xα and X−α do), yu y −1 ∈ U . Since |aα a | = |aα | + |a | > |aα | + |a | = 2 2 |aα a | = |a| , we have (∗), modulo the rank 1 case. This case, essentially G = SL2 , will be left as an exercise. (b) Assume x ∈ G, x = k1 ak2 as in (a). Then σx = k1 a−1 k2 , so that xσx−1 = 2 −1

(σa) = μ

(a)−1 = μ

(a−1 ) for all μ

∈ L.  k1 a k1 . Here σa = a−1 since μ (∗)

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8. VARIANTS OF THE BRUHAT LEMMA

67

Lemma 47. If elements of H are conjugate in G (any Chevalley group), they are conjugate under the Weyl group. Proof. This easily follows from the uniqueness in Theorem 4 .



By the lemma, x above uniquely determines a2 up to conjugacy under the Weyl group, hence also a since square-roots in A are unique. Remark. We can get uniqueness in (b) by replacing A by

(a) ≥ 1 for all α

> 0} . A+ = {a ∈ A : α This follows from Appendix III 33. Corollary. Let P consist of the elements of G which satisfy σx = x−1 and have all eigenvalues positive. (a) A ⊂ P . (b) Every p ∈ P is conjugate under K to some a ∈ A, uniquely determined up to conjugacy under W (spectral theorem). (c) G = KP , with uniqueness on the right (polar decomposition). Proof. (a) This has been noted in (b) above. (b) We can assume p = ka ∈ KA, by the theorem. Apply σ −1 : p = ak−1 . Thus k commutes with a2 , hence also with a. (Since a is diagonal (relative to a basis of weight vectors) and positive, the matrices commuting with a have a certain block structure which does not change when it is replaced by a2 .) Then k2 = 1 and k = a−1/2 pa−1/2 ∈ P , so that k is unipotent by the definition of P . Since K is compact, k = 1. Thus p = a. The uniqueness in (b) follows as before. (c) If x ∈ G, then x = k1 ak2 as in the theorem, so that x = k1 k2 ·k2−1 ak2 ∈ KP . Thus G = KP . Assume k1 p1 = k2 p2 with ki ∈ K and pi ∈ P . By (b) we can assume that p2 ∈ A. Then p1 = k1−1 k2 p2 . As in (b) we conclude that k1−1 k2 = 1, whence the uniqueness in (c).  Example. If G = SLn (C), so that K = SUn (C), A = {positive diagonal matrices} P = {positive-definite Hermitian matrices} then (b) and (c) reduce to classical results. We can now consider the case (c) of Lemma 43. The development is strikingly parallel to that for case (b) just completed although the results are basically arithmetic in one case, geometric in the other. Throughout we assume that θ, θ ∗ , k, Yα are as in Lemma 43(c) and that the Chevalley group G under discussion is based on k. We write Gθ for the subgroup of elements of G whose coordinates, relative to the original lattice in M , all lie in θ. Lemma 48. If ϕα is as in Lemma 4  , Corollary 6, then ϕα SL2 (θ) ⊆ Gθ . Proof. If θ is a Euclidean domain then SL2 (θ) is generated by its unipotent superdiagonal and subdiagonal elements, so that the lemma follows from the fact that xα (t) acts on M as an integral polynomial in t. In the general case it follows that if p is a prime in θ and θp is the localization of θ at#p (all a/b ∈ k such that a, b ∈ θ with b prime to p) then ϕα SL2 (θ) ⊆ Gθp . Since p θp = θ, e.g., by unique factorization, we have our result.  Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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8. VARIANTS OF THE BRUHAT LEMMA

A version of Lemma 48 is true if θ is any commutative ring since $Remark. % ϕα ac db is generically expressible as a polynomial in a, b, c, d with integral coefficients (proof # omitted). The proof just given works if θ is any integral domain for which θ = θp (p = maximal ideal), which includes most of the interesting cases. Lemma 49. Write K = Gθ , Kα = Gα ∩ K. (a) B ∩ K = (U ∩ K)(H ∩ K). (b) U ∩ K = { α>0 xα (tα ) : tα ∈ θ}. ¶ ©

= { hi (ti ) : ti ∈ θ ∗ }. (c) H ∩ K = h ∈ H : μ

(h) ∈ θ ∗ for all μ

∈L (d) ϕα SL2 (θ) = Kα . Hence Yα ⊂ Kα . Proof. (a) If b = uh ∈ B ∩ K, then its diagonal h, relative to a basis of M made up of weight vectors (see Lemma 18, Corollary 3), must be in K, hence u must also. (b) If u = xα (tα ) ∈ U ∩ K, then by induction on heights, the equation xα (t) = 1 + tXα + · · · and the primitivity of Xα in End(M ) (Theorem 2, Corollary 2) we get all tα ∈ θ. (c) If h ∈ H ∩ K, in diagonal form as above, then μ

(h) must be in θ for each weight μ

of the representation defining G, in fact in θ ∗ since the sum of these weights is 0 (the sum is invariant under W ). If we write h = hi (ti ) and use what has just been proved, we get tni ∈ θ ∗ for some n > 0, whence ti ∈ θ ∗ by unique factorization. (d) Set Sα = ϕα SL2 (θ). By Lemma 48, Sα ⊆ Kα . Since Gα = Bα ∪ Bα Yα by Lemma 43(c) and Yα ⊂ Sα , the reverse inclusion follows from: Bα ∩ K ⊂ Sα . Now if x = xα (t)hα (t∗ ) ∈ Bα ∩ K, then t ∈ θ and t∗ ∈ θ ∗ by (a), (b), (c) applied to Gα ,  so that x ∈ xα (θ), x−α (θ) = Sα , whence (d). Theorem 18. Let θ, k, G and K = Gθ be as above. Then G = BK (Iwasawa decomposition). Proof. By Lemma 43(c) and 49(d), BwB = BYα · · · Yδ ⊆ BK for every w ∈ W , so that G = BK.  Corollary 1. Write Kw = BwB ∩ K.  (a) K = w∈W Kw . (b) Kw = (B ∩ K)Yα · · · Yδ , with B ∩ K given by Lemma 49, and on the right there is uniqueness of expression. Remark. This normal form in K = Gθ has all components in Gθ whereas the usual one obtained by imbedding Gθ in G does not. Corollary 2. K is generated by the groups Kα . Proof. By Lemma 49 and Corollary 1.



Corollary 3. If θ is a Euclidean domain, then K is generated by {xα (t) : α ∈ Σ, t ∈ θ}. Proof. Since the corresponding result holds for SL2 (θ), this follows from Lemma 49(d) and Corollary 2.  Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

8. VARIANTS OF THE BRUHAT LEMMA

69

Example. Assume θ = Z, k = Q. We get that GZ is generated by {xα (1)}. The normal form in Corollary 1 can be used to extend Nielsen’s theorem (see (1) on page 58) from SL3 (Z) to GZ whenever Σ has rank ≥ 2, is indecomposable, and has all roots of equal length (see Wardlaw [66]). It would be nice if the form could be used to handle SL3 (Z) itself since Nielsen’s proof is quite involved. The case of unequal root lengths is at present in poor shape. In analogy with the fact that in the earlier development K is a simple compact group if Σ is indecomposable, we have here: Every normal subgroup of GZ is finite or of finite index if Σ is indecomposable and has rank ≥ 2. The proof isn’t easy. Exercise. Prove that GZ /DGZ is finite, and is trivial if Σ is indecomposable and not of type A1 , B2 or G2 . Returning to the general setup, if p is prime in θ, we write |·|p for the p-adic norm defined by |0|p = 0 and |x|p = 2−r if x = pr a/b with a and b prime to p. Theorem 19 (Approximation theorem). Let θ and k be as above, a principal ideal domain and its quotient field, S a finite set of inequivalent primes in θ, and for each p ∈ S, tp ∈ k. Then for any  > 0 there exists t ∈ k such that |t − tp |p <  for all p ∈ S and |t|q ≤ 1 for all primes q ∈ / S. Proof. We may assume every tp ∈ θ. To see this write tp = pr a/b as above. By choosing s ≥ −r and c and d so that a = cps + db and replacing a/b by d, we may assume b = 1. If we then multiply by a sufficiently high power of the product of the elements of S, we achieve r ≥ 0, for all p ∈ S. If we now choose n so that < , e = p∈S pn , ep = e/pn , then fp , gp so that fp pn + gp ep = 1, and finally 2−n   t = gp ep tp , we achieve the requirements of the theorem. Now given a matrix x = (aij ) over k, we define |x|p = max |aij |p . The following properties are easily verified. © ¶ (1) |x + y|p ≤ max |x|p , |y|p . (2) |xy|p ≤ |x|p |y|p . (3) If |xi |p = |yi |p for i = 1, 2, . . . , n, then & ¶ ©  && & yi & ≤ max |y1 |p · · · | yi |p · · · |yn |p |xi − yi |p . & xi − p

i

Theorem 20 (Approximation theorem for split groups). Let θ, k, S,  be as in Theorem 19, G a Chevalley group over k, and xp ∈ G for each p ∈ S. Then there / S. exists x ∈ G so that |x − xp |p <  for all p ∈ S and |x|q ≤ 1 for all q ∈ Proof. Assume first that all xp are contained ¶ in some © Xα , xp = xα (tp ) with tp ∈ k. If x = xα (t), t ∈ k, then |x|q ≤ max |t|q , 1 because xα (t) is an in& & & tegral polynomial in t and similarly &xx−1 p − 1 p ≤ |t − tp |p , so that |x − xp |p ≤ |xp |p |t − tp |p by (1) and (2) above. Thus our result follows from Theorem 19 in this case. In the general case we choose a sequence of roots α1 , α2 , . . . so that xp = xp1 xp2 · · · with xpi ∈ Xαi for all p ∈ S. By the first case there exists xi ∈ Xαi so that |xi − xpi |p < min(|xpi |p ,  |xpi |p / |xp1 |p |xp2 |p · · · ) if p ∈ S and |xi |q ≤ 1 if q ∈ / S. We set x = x1 x2 · · · . Then the conclusion of the theorem holds by (3) above.  Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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With Theorem 20 available we can now prove: Theorem 21 (Elementary divisor theorem). Assume θ, k, G, K = Gθ are as

(h) ∈ θ for all positive roots α

. before. Let A+ be the subset of H defined by: α + (a) G = KA K (Cartan decomposition). (b) The A+ component in (a) is uniquely determined mod H ∩ K, i.e., mod units (see Lemma 49); in other words, the set of numbers { μ(h) : μ

weight of the representation defining G} is uniquely determined. Example. The classical case occurs when G = SLn (k), K = SLn (θ), and A+ consists of the diagonal elements diag(a1 , a2 , . . . , an ) such that ai is a multiple of ai+1 for i = 1, 2, . . .. Proof of theorem. First we reduce the theorem to the local case, in which θ has a single prime, modulo units. Assume the result is true in this case. Assume x ∈ G. Let S be the finite set of primes at which x fails to be integral. For p ∈ S, we write θp for the local ring at p in θ, and define Kp and A+ p in terms of θp as K and A+ are defined for θ. By the local case of the theorem we may write x = cp ap cp with cp , cp ∈ Kp and ap ∈ A+

(ap ) p , for all p ∈ S. Since we may choose ap so that μ is always a power of p and then replace all ap by their product, adjusting the c’s accordingly, we may assume that ap is independent of p, is in A+ , and is integral outside of S. We have cp acp x−1 = 1 with a = ap for p ∈ S. By Theorem 20 there exist c, c ∈ G so that |c − cp |p < |cp |p for p ∈ S and |c|q ≤ 1 for q ∈ / S, & & the same equations hold for c and cp , and &cac x−1 − 1&p ≤ 1 for all p ∈ S. By & & properties (1), (2), (3) of |·|p , it is now easily verified that |c|p ≤ 1, &cp &p ≤ 1 and & &  −1 &cac x − 1& ≤ 1, whether p is in S or not. Thus c ∈ K, c ∈ K and cac x−1 ∈ K, p so that x ∈ KA+ K as required. The uniqueness in Theorem 21 clearly also follows from that in the local case. We now consider the local case, p being the unique prime in θ. The proof to follow is quite close to that of Theorem 17. Let A be the subgroup of all h ∈ H such that all μ

(h) are powers of p, and redefine A+ , casting out units, so that in addition all α

(h) ( α > 0) are nonnegative powers of p. Lemma 50. For each a ∈ A there exists a unique H ∈ HZ , the Z-module

(a) = pμ(H) for all generated by the elements Hα of the Lie algebra L, such that μ weights μ.

(a) = Proof. Write a = hα (cα pnα ) with cα ∈ θ ∗ , nα ∈ Z. Then we have μ (cα pnα )μ(Hα ) . Since μ

(a) is a power of p the cα , being units, may be omitted,  so that μ

(a) = pμ(H) with H = nα Hα . If H  is a second possibility for H, then    μ(H ) = μ(H) for all μ, so that H = H. If a and H are as above, we write H = logp a, a = pH , and introduce a norm: |a| = |H|, the Killing norm. This norm is invariant under the Weyl group. Now assume x ∈ G. We want to show x ∈ KA+ K. From the definitions if T = H ∩ K then H = AT . Thus by Theorem 18 there exists y = ua ∈ KxK with u ∈ U , a ∈ A. There is only a finite number of possibilities for a: if a = pH , then {μ(H) : μ a weight in the given representation} is bounded below (by −n if n is Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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71

  chosen so that the matrix of pn x is integral, because pμ(H) are the diagonal entries of y), and also above since the sum of the weights is 0, so that H is confined to a bounded region of the lattice HZ . We choose y = ua above so as to maximize |a|. If u = uα (uα ∈ Xα ), we set supp u = {α : uα = 1} and then minimize supp u subject to a lexicographic ordering of the supports based on an ordering of the roots consistent with addition (thus supp u < supp u means that the first α in one but not the other lies in the second). We claim u = 1. Suppose not. We claim (∗) uα ∈ / K and a−1 uα a ∈ / K for α ∈ supp u. If uα were not in K, we could move it to the extreme left in an expression for y and then remove it. The new terms introduced by this shift would, by the relations (B), correspond to roots higher than α, so that supp u would be diminished, a contradiction. Similarly a shift to the right yields the second part of (∗). Now as in the proof of Theorem 17 we may conjugate y by a product of wβ (1)’s (all in K) to get uα = 1 for some simple α, as well as (∗). We write a = pH , choose c so that  H  = H − cHα is orthogonal to Hα , set aα = pcHα , a = pH , a = aα a . We only know that 2c = H, Hα  ∈ Z, so that this may involve an adjunction of p1/2 which must eventually be removed. If we bear this in mind, then after reducing (∗) to the rank 1 case, exactly as in the proof of Theorem 17, what remains to be proved is this: Lemma 51. Assume26

ï

y = ua =

1 t 0 1

òï

pc

ò p−c

with 2c ∈ Z, t ∈ k, t ∈ / θ and tp−2c ∈ / θ. Then c can be increased by an integer by multiplications by elements of K. Proof. Let t = ep−n with e ∈ θ ∗ . Then n ∈ Z, n > 0 and n + 2c > 0 by the assumptions, so that c + n > c, c + n > −c and |c + n| > |c|. If we multiply y on the left by ò ï pn e −e−1 0 on the right by ï ò 1 0 −1 n+2c −e p 1 both in K, we get ï c+n ò p 0 0 p−c−n which proves the lemma.  Now u = 1. Thus y = a ∈ A, so that x ∈ KAK. Thus G = KAK. Finally every element of A is conjugate to an element of A+ under the Weyl group, which is fully represented in K (every wα (1) ∈ K). Thus G = KA+ K. It remains to prove the uniqueness of the A+ component. If G is the universal group of the same type as G and π is the natural homomorphism, it follows from Lemma 49(d) and Theorem 18, Corollary 2 that πK  = K and from Lemma 49 that π maps A+ 26 If

c∈ / Z, this notation means that we consider a representation SL2 such that

maps to a matrix with integer entries. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

'

pc 0

0

p−c

(

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8. VARIANTS OF THE BRUHAT LEMMA

isomorphically onto A+ . Thus we may assume that G is universal. Then G is a direct product of its indecomposable factors so that we may also assume that G is indecomposable. Let λi be the i-th fundamental weight, Vi an L-module with λi as highest weight, Gi the corresponding Chevalley group, πi : G → Gi the corresponding homomorphism, and μi the corresponding lowest weight. Assume “i (a) = p−ni . Each weight now that x = cac ∈ G, with c, c ∈ K and a ∈ A+ . Set μ on Vi is μi increased by a sum of positive roots. Thus ni is the smallest integer such that pni πi a is integral, i.e., such that pni πi x is since πi c and πi c are integral, thus is uniquely determined by x. Since {μi } is a basis of the lattice of weights (μi = wo λi ), this yields the uniqueness in the local case and completes the proof of Theorem 21.  Corollary 1. If θ is not a field, the group K is maximal in its commensurability class.27 Proof. Assume K  is a subgroup of G containing K properly. By the theorem / K. Some entry of the diagonal matrix a is nonintegral there exists a ∈ A+ ∩K  , a ∈ so that by unique factorization |K  /K| is infinite.  Remark. The case θ = Z is of some importance here. Corollary 2. If θ = Zp and k = Qp (p-adic integers and numbers) and the p-adic topology is used, then K is a maximal compact subgroup of G. Proof. We will use the fact that Zp is compact. (The proof is a good exercise.) We may assume that G is universal. Let k be the algebraic closure of k and G the corresponding Chevalley group. Then G = G ∩ SL(V, k) (Theorem 7, Corollary 3), so that K = G ∩ SL(V, θ). Since θ is compact, so is End(V, θ), hence also is K, the set of solutions of a system of polynomial equations since G is an algebraic group, by Theorem 6. If K  is a subgroup of¶G containing©K properly, there exists a ∈ A+ ∩ K  , a ∈ / K, by the theorem. Then |an |p : n ∈ Z is not bounded so that  K is not compact.  Remark. We observe that in this case the decompositions G = BK and G = KA+ K are relative to a maximal compact subgroup just as in Theorems 16 and 17. Also in this case the closure formula of Theorem 16, Corollary 7 holds. Exercise (Optional). Assume that G is a Chevalley group over C, R, or Qp and that K is the corresponding maximal compact subgroup discussed above. Prove the commutativity under convolution of the algebra of functions on G which are complex-valued, continuous, with compact support, and invariant under left and right multiplications by elements of K. (Such functions are sometimes called zonal functions and are of importance in the analysis of G.) Hint: prove that there exists an antiautomorphism ϕ of G such that ϕxα (t) = x−α (t) for all α and t, that ϕ preserves every double coset relative to K, and that ϕ preserves Haar measure. A much harder exercise is to determine the exact structure of the algebra. Next we consider a double coset decomposition of K = Gθ itself in the local case. We will use the following result, the first step in the proof of Theorem 7.28 27 Two subgroups are called commensurable if their intersection has finite index in each of them. () 28 See Macdonald [42]. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Lemma 52. Let L be the Lie algebra of G (the original Lie algebra of Chapter 1 with its coefficients transferred to k), N the number of positive roots, and " the exterior power N L made up of {Y1 , Y2 , . . . , Yr } a basis of " products of the cj (x)Yj . Then Xα ’s and Hi ’s with Y1 = α>0 Xα . For x ∈ G write xY1 = x ∈ U − HU if and only if c1 (x) = 0. Theorem 22. Assume that θ is a local principal ideal domain, that p is its unique prime, and that k and G are as before. (a) BI = Up− Hθ Uθ is a subgroup of Gθ .  (b) Gθ = w∈W BI wBI (disjoint), if the representatives for W in G are chosen in Gθ . (c) BI wBI = BI wUw,θ with the last component of the right uniquely determined mod Uw,p . Proof. Let θ¯ denote the residue class field θ/pθ, Gθ¯ the Chevalley group of ¯ and B ¯, H ¯, . . . the usual subgroups. By Theorem 18, the same type as G over θ, θ θ Corollary 3 reduction mod p yields a homomorphism π of Gθ onto Gθ¯.   " (1) π −1 Uθ¯− Hθ¯Uθ¯ ⊆ U − HU . We consider G acting on N L as in Lemma 52. As is easily seen Gθ acts integrally relative to the basis of Y ’s. Now assume πx ∈ Uθ¯− Hθ¯Uθ¯. Then c1 (πx) = 0 by the lemma applied to Gθ¯, whence c1 (x) = 0 and x ∈ U − HU again by the lemma. (2) Corollary. ker π ⊆ U − HU . (3) BI = π −1 Bθ¯. Assume x ∈ π −1 Bθ¯. Then x ∈ U − BU by (1). From this and x ∈ Gθ it follows as in the proof of Theorem 7(b) that x ∈ Uθ− Hθ Uθ , and then that x ∈ BI . (4) Completion of proof: By (3) we have (a). To get (b) we simply apply π −1 to the decomposition in Gθ¯ relative to Bθ¯. We need only remark that a choice as indicated is always possible since each wα (1) ∈ Gθ . From (b) the equation in (c) easily follows. (Check this.) Assume b1 wu1 = b2 wu2 with bi ∈ BI , ui ∈ Uw,θ . Then −1 −1 b−1 ∈ BI ∩ Uθ− = Up− , whence u1 u−1  1 b2 = wu1 u2 w 2 ∈ Uw,p and (c) follows. Remark. The subgroup BI above is called an Iwahori subgroup. It was introduced in an interesting paper by Iwahori and Matsumoto [33]. There a decomposition which combines those of Theorems 21(a) and 22(b) can be found. The present development is completely different from theirs.  There is an interesting connection between the decomposition Gθ = BI wBI above and the one, Gθ = (BwB)θ , that Gθ inherits as a subgroup of G, namely: Corollary. Assume w ∈ W , that S(w) is as in Theorem 16, Corollary 9, and that π : Gθ → Gθ¯is, as above, the natural projection. Then π(BI wBI ) = Bθ¯wBθ¯, and π(BwB)θ = w ∈S(w) Bθ¯w Bθ¯. Hence if θ¯ is a topological field, e.g., C, R or Qp , then π(BwB)θ is the topological closure of π(BI wBI ). Proof. The first equation follows from π −1 Bθ¯ = BI , proved above. Write w = wα wβ · · · as in Lemma 25, Corollary. Then (∗) (BwB)θ = (Bwα B)θ (Bwβ B)θ · · · by Theorem 18, Corollary 1. Now x−α (p) and wα (1) are elements of (Bwα B)θ , and (Bwα B)θ is a union of Bθ double cosets. Thus, π(Bwα B)θ ⊇ Bθ¯ ∪ Bθ¯wα Bθ¯ = Bθ Gα,θ¯. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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The reverse inequality also holds since (Bwα B)θ ⊆ Bθ Gα,θ by Theroem 18, Corollary 1. From this, (∗), the definition of S(w), and Lemma 25, the required expression  for π(BwB)θ now follows. Appendix. Our purpose is to prove Theorem 23 below which gives the closure of BwB under very general conditions. We will write w ≤ w if w ∈ S(w) with S(w) as in Theorem 16, Corollary 9, i.e., if w is a subexpression (i.e., a product of a subsequence) of some minimal expression of w as a product of simple reflections. Lemma 53. The following are true. (a) If w is a subexpression of some minimal expression for w, it is a subexpression of all of them. (b) In (a) the subexpression for w can be taken to be minimal. (c) The relation ≤ is transitive. (d) If w ∈ W and α is a simple root such that wα > 0 (resp. w−1 α > 0), then wwα > w (resp. wα w > w). (e) w0 ≥ w for all w ∈ W . Proof. (a) This was proved in Theorem 16, Corollary 7 and 9 in a rather roundabout way. It is a direct consequence of the following fact, which will be proved in a later section: the equality of two minimal expressions for w (as a product of simple reflections) is a consequence of the relations w1 w2 · · · = w2 w1 · · · (w1 , w2 distinct simple reflections, n terms on each side, n = order of w1 w2 ). (b) If w = w1 w2 · · · wr is an expression as in (b) and it is not minimal, then two of the terms on the right can be canceled by Appendix II 21. (c) By (a) and (b). (d) If wα > 0 and w1 w2 · · · ws is a minimal expression for w, then w1 · · · ws wα is one for wwα by Appendix II 19, so that wwα > w, and similarly for the other case. (e) This is proved in Lemma 46.  Now we come to our main result. Theorem 23. Let G be a Chevalley group. Assume that k is a nondiscrete topological field and that the topology inherited by G as a matrix group over k is used (Zariski or from k). Then the following conditions on w, w are equivalent. (a) Bw B ⊆ BwB. (b) w ≤ w. " Proof. Let Y1 be as in Lemma 52 and more generally Yw = α>0 Xwα for w ∈ W . For x ∈ G let cw (x) denote the coefficient of Yw in xY1 . We will show that (a) and (b) are equivalent to: (c) cw is not identically zero on BwB.  (a) ⇒ (c). We have xβ (t)Xα = Xα + tj Xj with Xj of weight (0 or a root) α + jβ, and nw Xα = cXwα (c = 0) if nw represents w in W in N/H. Thus (∗) BwBY1 ⊆ k∗ Yw + higher terms in the ordering given by sums of positive roots. Thus cw is not identically zero on Bw B, hence also not on BwB, by (a). (c) ⇒ (b). We use downward induction on N (w ). If this is maximal then  w = w0 , the element of W making all positive roots negative, and then w = w0 by Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(c) and (∗) above. Assume w = w0 . Choose α simple so that w−1 α > 0, hence N (wα w ) > N (w ). Since cw (BwB) = 0 and Bwα bwB ⊆ BwB ∪ Bwα wB, we see that cwα w (BwB) = 0 or cwα w (Bwα wB) = 0, so that wα w ≤ w or wα w ≤ wα w. In the first case w < w by Lemma 53(c) and (d). In the second case if w−1 α < 0 then wα w < w by Lemma 53(d) which puts us back in the first case, while if not we may choose a minimal expression for w starting with wα and conclude that w ≤ w. (b) ⇒ (a). By the definitions and the usual calculations of double cosets, this is equivalent to: if α is simple, then Bwα B = B ∪ Bwα B. The left side is contained in the right, an algebraic group, hence a closed subset of G. Since Bwα B contains Xα − {1} and the topology on k is not discrete, its closure contains 1, hence also B, proving the reverse inequality and completing the proof of the theorem.  Remark. In case k above is C, R or Qp , the theorem reduces to results obtained earlier. In case k is infinite and the Zariski topology on k and G are used it becomes a result of Chevalley29 (unpublished). Our proof is quite different from his. Exercise. (a) If w ∈ W and α is a positive root such that wα > 0, prove that wwα > w (compare this with Lemma 53(d)), and conversely if w ≤ w then (∗) there exists a sequence of positive roots α1 , α2 , . . . , αr such that if wi = wαi then w w1 · · · wi−1 αi > 0 for all i and w w1 · · · wr = w. Thus conditions w ≤ w and (∗) are equivalent. (b) It seems to us likely that w ≤ w is also equivalent to: there exists a permutation π of the positiveroots such that w πα − wα is a sum of positive roots for every α > 0; or even to: α>0 (w α − wα) is a sum of positive roots. (c) Condition w ≤ w is equivalent to: w makes negative all the positive roots that w makes negative.

this is not true since the Zariski topology on the n-dimensional space kn does not coincide with the product of Zariski topologies on k. However, the proof can be adapted to this case without any changes. () 29 Formally,

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https://doi.org/10.1090//ulect/066/10

CHAPTER 9

The Orders of the Finite Chevalley Groups Presently we will prove: Theorem 24. Let W be a finite reflection group on a real space V of dimension l, S the algebra of polynomials on V , I(S) the subalgebra of invariants under W . Then: (a) I(S) is generated by l homogeneous algebraically independent elements I 1 , . . . , Il . (b) The  degrees of the Ij ’s, say d1 , . . . , dl , are uniquely determined and satisfy j (dj − 1) = N , the number of positive roots. (c) For the irreducible Weyl groups the di ’s are as follows: W Al Bl , C l Dl E6 E7 E8 F4 G2

di 2, 3, . . . , l + 1 2, 4, . . . , 2l 2, 4, . . . , 2l − 2, l 2, 5, 6, 8, 9, 12 2, 6, 8, 10, 12, 14, 18 2, 8, 12, 14, 18, 20, 24, 30 2, 6, 8, 12 2, 6

Our main goal is: Theorem 25. (a) Let G be a universal Chevalley group over a field k of q elements  and the di ’s as in Theorem 24. Then |G| = q N i (q di − 1) with N = (di − 1) the number of positive roots. (b) If G is simple instead, then we have to divide by c = |Hom(L1 /L0 , k∗ )|, given as follows:30 G Al Bl , C l Dl E6 E7 E8 c (l + 1, q − 1) (2, q − 1) (4, q l − 1) (3, q − 1) (2, q − 1) 1

F4 1

G2 1

Remark. We see that the groups of type Bl and Cl have the same order. If l = 2 the root systems are isomorphic so the groups are isomorphic. We will show later that if l ≥ 3 the groups are isomorphic if and only if q is even. 30 (a, b)

denotes the greatest common divisor of integers a and b. ()

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9. THE ORDERS OF THE FINITE CHEVALLEY GROUPS

The proof of Theorem 25 depends on the following identity. Theorem 26. Let W and the di ’s be as in Theorem 24 and t an indeterminate.  Then w∈W tN (w) = i (1 − tdi )/(1 − t). We first show that Theorem 25 is a consequence of Theorems 24 and 26.  Lemma 54. If G is as in Theorem 25(a) then |G| = q N (q − 1)l w∈W q N (w) .  Proof. Recall that, by Theorems 4 and 4 , G = w∈W BwB (disjoint) and  BwB = U HwUw with uniqueness of expression. Hence |G| = |U | |H| · w∈W |Uw |. Now by Corollary 1 to the proposition of Chapter 3, |U | = q N and |Uw | = q N (w) .  By Lemma 28, |H| = (q − 1)l . Corollary. U is a p-Sylow subgroup of G, if p denotes the characteristic of k. Proof. p | q N (w) unless N (w) = 0. Since N (w) = 0 if and only if w = 1,  p  w∈W q N (w) .  Proof of Theorem 25. (a) follows from Lemma 54 and Theorem 26. (b) follows from the fact that the center of the universal group is isomorphic to Hom(L1 /L0 , k∗ ) and the values of L1 /L0 found in Chapter 3.  Before giving general proofs of Theorem 24 and 26 we give independent (case by case) verifications of Theorems 24 and 26 for the classical groups. Theorem 24. Al : Here W∼ = Sl+1 permuting l + 1 linear functions ω1 , . . . , ωl+1 such that σ1 = ωi = 0. In this case the elementary symmetric polynomials σ2 , . . . , σl+1 are invariant and generate all other polynomials invariant under W . Bl , Cl : Here W acts relative to a suitable basis ω1 , . . . , ωl by all permutations and sign changes. Here the elementary symmetric polynomials in ω12 , . . . , ωl2 are invariant and generate all other polynomials invariant under W . Dl : Here only an even number of sign changes can occur. Thus we can replace the last of the invariants for Bl , ω12 · · · ωl2 by ω1 · · · ωl . Theorem 26. Al : Here W ∼ of inversions in the sequence = Sl+1 and N (w) is the number  N (w) (w(1), . . . , w(l + 1)). If we write Pl (t) = w∈W ∼ then Pl+1 (t) = =Sl+1 t 2 l+1 Pl (t)(1+t+t +· · ·+t ) as we see by considering separately the l+2 values j that w(l + 2) can take on. Hence the formula Pl (t) = l+1 j=2 (1 − t )/(1 − t) follows by induction. Exercise. Prove the corresponding formulas for types Bl , Cl and Dl . Here the proof is similar, the induction step being a bit more complicated. Part (a) of Theorem 24 follows from: Theorem 27. Let G be a finite group of automorphisms of a real vector space V of finite dimension l and I the algebra of polynomials on V invariant under G. Then: (a) If G is generated by reflections, then I is generated by l algebraically independent homogeneous elements (and 1). Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(b) Conversely, if I is generated by l algebraically independent homogeneous elements (and 1) then G is generated by reflections. Example. Let l = 2 and V have coordinates x, y. If G = {± id}, then G is not a reflection group. I is generated by x2 , xy and y 2 and no smaller number of elements suffices. Notation. Throughout the proof we let S be the algebra of all polynomials on V , S0 the ideal in S generated by the homogeneous elements of I of positive  degree, and Av stands for average over G (i.e., Av P = |G|−1 g∈G gP ). Proof of (a). (Chevalley [15].) (1) Assume I1 , I2 , . . . are elements of I such that I1 is not in the ideal in I generated  by the others and that P1 , P2 , . . . are homogeneous elements of S such that Pi Ii = 0. Then P1 ∈ S0 .  by I2 , . . .. Then I1 =  Proof of (1). Suppose I1 ∈ ideal in S generated R I for some R , . . . ∈ S so that I = Av I = 2 1 1 i≥2 i i i≥2 (Av Ri )Ii belongs to the ideal in I generated by I2 , . . ., a contradiction. Hence I1 does not belong to the ideal in S generated by I2 , . . .. Now we prove (1) by induction on d = deg P1 . If d = 0, P1 = 0 ∈ S0 . Assume d > 0 and let g ∈ G be  a reflection in a hyperplane L = 0. Then for each i, ((Pi − gPi )/L) Ii = 0, so by the induction assumption L | (Pi − gPi ). Hence P1 − gP1 ∈ S0 , i.e., P1 ≡ gP1 mod S0 . Since G is generated by reflections this holds for all g ∈ G and hence P1 ≡ Av P1 mod S0 . But Av P1 ∈ S0 so P1 ∈ S0 .  We choose a minimal finite basis I1 , . . . , In for S0 formed of homogeneous elements of I. Such a basis exists by Hilbert’s Theorem. (2) The Ii ’s are algebraically independent. Proof of (2). If the Ii are not algebraically independent, let H(I1 , . . . , In ) = 0 be a nontrivial relation with all monomials in the Ii ’s of the same degree in the underlying coordinates x1 , . . . , xl . Let Hi = ∂I∂ i H(I1 , . . . , In ). By the choice of H not all Hi are 0. Choose the notation so that {H1 , . . . , Hm } (m ≤ n) but no subset of it generates the ideal in I generated by all the Hi . Let Hj = m i=1 Vj,i Hi for j = m + 1, . . . , n where Vj,i ∈ I and all terms in the equation are homogeneous of the same degree. Then for k = 1, 2, . . . , l we have n  ∂Ii ∂H = Hi 0= ∂xk ∂xk i=1 ) * m n   ∂Ij ∂Ii . = Hi + Vj,i ∂xk j=m+1 ∂xk i=1  By (1) ∂x∂ k I1 + nj=m+1 Vj,i ∂x∂ k Ij ∈ S0 . Multiplying by xk , and summing over k,  using Euler’s formula31 , and writing dj = deg Ij we get d1 I1 + nj=m+1 Vj,1 dj Ij = n i=1 Ai Ii where Ai belongs to the ideal in S generated by the xk . By homogeneity  A1 = 0. Thus I1 is in the ideal generated by I2 , . . . , In , a contradiction. (3) The Ii ’s generate I as an algebra. 31

n i=1

∂f xi ∂x = (deg f )f . i

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 Proof of (3). Assume P ∈ I is homoegeneous of positive degree. Then P = Pi Ii , Pi ∈ S. By averaging we can assume that each Pi ∈ I. Each Pi is of degree less than or equal to the degree of P , so by induction on its degree P is a  polynomial in the Ii ’s. (4) n = l. Proof of (4). By (2) n ≤ l. By Galois theory R(I) is of finite index in  R(x1 , x2 , . . . , xn ), hence has transcendence degree l over R, whence n ≥ l. 

By (2), (3) and (4), (a) holds.

Proof of (b). (Shephard and Todd, [53].) Let I1 , . . . , Il be algebraically independent generators of I of degrees d1 , . . . , dl , respectively. (5) li=1 (1 − tdi )−1 = Avg∈G det(1 − gt)−1 , as a formal identity in t. Proof of (5). Let 1 , . . . , l be the eigenvalues of g and x1 , . . . , xl the corresponding eigenfunctions. Then det(1 − gt)−1 = i (1 + i t + 2i t2 + · · · ). The coefficient of tn is  p11 p22 · · · p1 +p2 +···=n

i.e., the trace of g acting on the space of homogeneous polynomials in x1 , . . . , xl of degree n, since the monomials xp11 xp22 · · · form a basis for this space. By averaging we get the dimension of the space of invariant homogeneous polynomials of degree n. This dimension is the number of monomials I1p1 I2p2 · · · of degree n, i.e., the number of solutions of p1 d1 +p2 d2 +· · · = n, i.e., the coeffiicient of tn in li=1 (1−tdi )−1 .   (6) di = |G| and (di − 1) = N = number of reflections in G. Proof of (6). We have ⎧ l ⎪ ⎨(1 − t) det(1 − gt) = (1 − t)l−1 (1 + t) ⎪ ⎩ a polynomial not divisible by (1 − t)l−1

if g = 1, if g is a reflection, otherwise.

Substituting this in (5) and multiplying by (1 − t)l , we have ã Å  −1 1 N (1 − t) + (1 − t2 )P (t) = 1 + t + · · · + tdi −1 1+ |G| (1 + t) where P (t) is regular at t = 1. Setting t = 1 we get d−1 = |G|−1 . Differentiating i and setting t = 1 we get Ä ä −1 d−1 (−(di − 1)/2) = |G| (−N/2) i  so (di − 1) = N .  (7) Let G be the subgroup of G generated by its reflections. Then G = G and hence G is a reflection group. Proof of (7). Let Ii , di and N  refer to G . The Ii can be expressed as polynomials in the Ii with the determinant of the corresponding Jacobian not 0. Hence after a rearrangement of the Ii , ∂I∂  Ii = 0 for all i. Hence di ≥ di . But i     (di − 1) = N = N = (d − 1) by (6). Hence di = di for all i, so, again by (6), i     |G| = di = di = |G |, so G = G . Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

9. THE ORDERS OF THE FINITE CHEVALLEY GROUPS

81

Corollary. The degrees d1 , d2 , . . . above are uniquely determined and satisfy (6). Thus Theorem 24(b) holds.



Exercise. For each positive root α, let lα ∈ S be a linear polynomial (unique up to multiplication by a nonzero scalar) determining the reflecting hyperplane of sα . Then Å ã  ∂Ii lα det = ∂xj α>0 up to multiplication by a nonzero number. Remark. The theorem remains true if R is replaced by any field of characteristic 0 and “reflection” is replaced by “automorphism of V with fixed point set a hyperplane.”32 For the proof of Theorem 24(c) (determination of the di ) we use: Proposition. Let G and the di be as in Theorem 27 and w = w1 · · · wl , the product of the simple reflections (relative to an ordering of V (see Appendix I.8)) in any fixed order. Let h be the order of w. Then: (a) N = lh/2. (b) w contains ω = exp 2πi/h as an eigenvalue, but not 1. (c) If the eigenvalues of w are {ω mi : 1 ≤ mi ≤ h − 1} then {mi + 1} = {di }. Proof. This was first proved by Coxeter [20], case by case, using the classification theory. For a proof not using the classification theory see Steinberg [58] for (a) and (b) and Coleman [19] for (c) using (a) and (b).33 This can be used to determine the di for all the Chevalley groups. As an example consider the di for E8 . Here l = 8, N = 120, so by (a) h = 30. Since w acts rationally {ω n : (n, 30) = 1} are all eigenvalues. Since ϕ(30) = 8 = l these are all the eigenvalues. Hence the di are 1, 7, 11, 13, 17, 19, 23, 29 all increased by 1, as listed previously. The proofs for G2 and F4 are exactly the same. E6 and E7 require further argument.  Exercise. Argue further. Remark. The di ’s also enter into the following results, related to Theorem 24: (a) Let L be the original Lie algebra, k a field of characteristic 0, G the corresponding adjoint Chevalley group. The algebra of polynomials on L invariant under G is generated by l algebraically independent alemenets of degree d1 , . . . , dl , the di ’s as above. This is proved by showing that under restriction from L to H the G-invariant polynomials on L are mapped isomorphically onto the W -invariant polynomials on H. The corresponding result for the universal enveloping algebra of L then follows easily. (b) If G acts on the exterior algebra on L, the algebra of invariants is an exterior algebra generated by l independent homogeneous elements of degrees {2di − 1}. 32 The 33 See

groups obtained using this construction are described by Shephard and Todd [53]. () also Bourbaki [9, Chapter V, § 6]. ()

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This is more difficult. It implies that the Poincar´e polynomial (whose coefficients are the Betti numbers) of the corresponding compact semisimple Lie group (the group K constructed from C in Chapter 8) is (1 + t2di −1 ). Proof of Theorem 26. (Solomon [54].) Let Π be the set of simple roots. If π ⊆ Π let Wπ be the subgroup generated by all wα , α ∈ π. (1) If w ∈ Wπ then w permutes the positive roots with support not in π.  Proof of (1). If β is a positive root and supp β  π then β = α∈Π θα α / π. Now wβ is β plus a vector with support in π, hence its with some θα > 0, α ∈ coefficient of α is positive, so wβ > 0.  (2) Corollary. If w ∈ Wπ then N (w) is unambiguous (i.e., it is the same whether we consider w ∈ W or w ∈ Wπ ). (3) For π ⊆ Π define Wπ = {w ∈ W : wπ > 0}. Then: (a) Every w ∈ W can be written uniquely as w = w w with w ∈ Wπ and w ∈ Wπ . (b) In (a) N (w) = N (w ) + N (w ). Proof of (3). (a) For any w ∈ W let w ∈ Wπ w be such that N (w ) is minimal. Then w α > 0 for all α ∈ π by Appendix II.19(a ). Hence w ∈ Wπ so that w ∈ Wπ Wπ . Suppose now w = w w = u u with w , u ∈ Wπ and w , u ∈ Wπ . Then w w u−1 = u . Hence w w u−1 π > 0. Now w (−π) < 0 so w u−1 π has support in π. Hence w u−1 π ⊆ π so by Appendix II.23 (applied to Wπ ) w u−1 = 1. Hence w = u , w = u . (b) This follows from (a) and (1).    (4) Let W (t) = w∈W tN (w) , Wπ (t) = w∈Wπ tN (w) . Then  (−1)π W (t)/Wπ (t) = tN π⊆Π

where N is the number of positive roots and (−1)π = (−1)|π| .  N (w) . Therefore Proof of (4). We have, by (3), W (t)/Wπ (t) = w∈Wπ t N (w) t where cw = the contribution of the term for w to the sum in (4) is c w  π (−1) . If w keeps positive exactly k elements of Π then π⊂Π,wπ>0 ® (1 − 1)k = 0, if k = 0, cw = 1, if k = 0. Therefore the only contribution is made by w0 , the element of w which makes all  positive roots negative, so the sum in (4) is equal to tN as required.  Corollary. (−1)π |W | / |Wπ | = 1. Exercise. Deduce from (4) that if α and β are complementary subsets of Π then   (−1)π−α /Wπ (t) = (−1)π−β /Wπ (t−1 ). π⊇α

π⊇β

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9. THE ORDERS OF THE FINITE CHEVALLEY GROUPS

83

Set D = {v ∈ V : (v, α) ≥ 0 for all α ∈ Π}, and for each π ⊆ Π set Dπ = {v ∈ V : (v, α) = 0 for all α ∈ π, (v, β) > 0 for all β ∈ Π − π} . Dπ is an open face of D. (5) The following subgroups of W are equal: (a) (b) (c) (d)

Wπ . The stabilizer of Dπ . The point stabilizer of Dπ . The stabilizer of any point of Dπ .

Proof of (5). (a) ⊆ (b) because π is orthogonal to Dπ . (b) ⊆ (c) because D is a fundamental domain for W by Appendix III.33. Clearly (c) ⊆ (d). (d) ⊆ (a) by Appendix III.32.  (6) In the complex cut on real  k-space by a finite number of hyperplanes let ni be the number of i-cells. Then (−1)i ni = (−1)k . Proof of (6). This follows from Euler’s formula, but may be proved directly by induction. In fact, if an extra hyperplane H is added to the configuration, each original i-cell cut in two by H has corresponding to it in H an (i−1)-cell separating   the two parts from each other, so that (−1)i ni remains unchanged. (7) In the complex K cut from V by the reflecting hyperplanes let nπ (w) (π ⊆  Π, w ∈π W ) denote the number of cells W -congruent to Dπ and w-fixed. Then π⊆Π (−1) nπ (w) = det w. Proof of (7). Each cell of K is W -congruent to exactly one Dπ . By (5) every (Vw = {v ∈ V : wv = v}). Applying (6) to Vw and using cell fixed by w lies in Vw  dim Dπ = l − |π| we get π⊆Π (−1)π nπ (w) = (−1)l−k , where k = dim Vw . But w is orthogonal, so that its possible eigenvalues in V are +1, −1 and pairs of complex  conjugate numbers. Hence (−1)l−k = det w. If χ is a character on W1 , a subgroup of W , then χW denotes the induced character defined by  1 χ(xwx−1 ). (∗) χW (w) = |W1 | w∈W xwx−1 ∈W1

(See, e.g., Feit [25].) (8) Let χ be a character on W and χπ = (χ|Wπ )W (π ⊆ Π). Then  (−1)π χπ (w) = χ(w) det w π⊆Π

for all w ∈ W . Proof of (8). Assume first that χ ≡ 1. Now xwx−1 ∈ Wπ if and only if xwx−1 fixes Dπ (by (5)) which happens if and only if w fixes x−1 Dπ . Therefore 1π (w) = nπ (w) by (∗). By (7) this gives the result for χ ≡ 1. If χ is any character  then χπ = χ · 1π so (8) holds. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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9. THE ORDERS OF THE FINITE CHEVALLEY GROUPS

(9) Let M be a finite-dimensional real W -module, Iπ (M ) be the subspace of

) be the space of W -skew-invariants, i.e., Wπ -invariant, and I(M

) = {m ∈ M : wm = (det w)m for all w ∈ W } . I(M 

). Then π⊆Π (−1)π dim Iπ (M ) = dim I(M Proof of (9). In (8) take χ to be the character of M , average over w ∈ W , and use (∗).  (10) If p = α>0 lα , where lα is a linear polynomial (unique up to multiplication by a nonzero scalar) determining the reflecting hyperplane of the reflection wα , then p is skew and p divides every skew polynomial on V . Proof of (10). We have wα p = −p = (det wα )p if α is a simple root by Appendix I.11. Since W is generated by simple reflections p is skew. If f is skew and α is a root then wα f = (det wα )f = −f so α | f . By unique factorization p | f.  di (11) Let P (t) = (1 − t )/(1 − t) and  for π ⊆ Π let {dπi } and Pπ be defined for Wπ as {di } and P are for W . Then π⊆Π (−1)π P (t)/Pπ (t) = tN . Proof of (11). We must show    (∗) (−1)π (1 − tdπi )−1 = tN (1 − tdi )−1 . ∞

π⊆Π

i

i

Let S = k=0 Sk be the algebra of polynomials on V , graded as usual. As in (5)of the proof of Theorem 27 the coefficient of tk on the left hand side of (∗) is π⊆Π (−1)π dim Iπ (Sk ). Similarly, using (10), the coefficient of tk on the right

k ). These are equal by (9). hand side of (∗) is dim I(S  (12) Proof of Theorem 26. We write (11) as  N  t − (−1)Π /P (t) = (−1)π /Pπ (t) π⊆Π

   and (4) as tN − (−1)Π /W (t) = π⊆Π (−1)π /Wπ (t). Then, by induction on |Π|, W (t) = P (t).  Remark. Step (7), the geometric step, represents the only simplification of Solomon’s original proof.

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https://doi.org/10.1090//ulect/066/11

CHAPTER 10

Isomorphisms and Automorphisms In this chapter we discuss the isomorphisms and automorphisms of Chevalley groups over perfect fields. This assumption of perfectness is not strictly necessary but it simplifies the discussion in one or two places. We begin by proving the existence of certain automorphisms related to the existence of symmetries of the underlying root systems. Lemma 55. Let Σ be an abstract indecomposable root system with not all roots of one length. Let Σ∗ = {α∗ = 2α/(α, α) : α ∈ Σ} be the abstract system obtained by inversion. Then: (a) Σ∗ is a root system. (b) Under the map ∗ long roots are mapped onto short roots and vice versa. Further, angles and simple systems of roots are preserved. (c) If p = (α0 , α0 )/(β0 , β0 ) with α0 long, β0 short then the map ® α →

pα∗ , α∗ ,

if α is long, if α is short,

extends to a homothety. Proof. (a) holds since α∗ , β ∗  = β, α. (b) and (c) are clear.



The root system Σ∗ obtained in this way from Σ is called the root system dual to Σ.  Exercise. Let α = ni αi be a root expressed in terms of the simple ones. Prove that α is long if and only if p | ni whenever αi is short. Examples. (a) For n ≥ 3, Bn and Cn are dual to each other. B2 and F4 are in duality with themselves (with p = 2) as is G2 (with p = 3). (b) Let α, β, α + β, α + 2β be the positive roots for Σ of type B2 . Then those for Σ∗ are α∗ , β ∗ , (α + β)∗ = 2α∗ + β ∗ , and (α + 2β)∗ = α∗ + β ∗ . If we identify α∗ with β and β ∗ with α we get a map of B2 onto itself. α → β, β → α, α + β → α + 2β, α + 2β → α + β. This is the map given by reflecting in the line L in the diagram below (L is the bisector of ∠(α, β)) and adjusting lengths. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 85

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L α+β α + 2β α β Theorem 28. Let Σ, Σ∗ and p be as above, k a field of characteristic p (p is either 2 or 3), G, G∗ universal Chevalley groups constructed from (Σ, k) and (Σ∗ , k) respectively. Then there exists a homomorphism ϕ of G into G∗ and signs α for all α ∈ Σ such that ® xα∗ (α t), if α is long, ϕ(xα (t)) = xα∗ (α tp ), if α is short. If k is perfect then ϕ is an isomorphism of abstract groups. Examples. (a) If k is perfect of characteristic 2 then Spin2n+1 , SO2n+1 (split forms), and Sp2n are isomorphic. (b) Consider C2 , p = 2, α = 1. The theorem asserts that on U we have an endomorphism (as before we identify Σ and Σ∗ ) such that (1) ϕ(xα (t)) = xβ (t), ϕ(xβ (t)) = xα (t2 ), ϕ(xα+β (t)) = xα+2β (t2 ), ϕ(xα+2β (t)) = xα+β (t). The only nontrivial relation of type (B) on U is (2) (xα (t), xβ (u)) = xα+β (tu)xα+2β (tu2 ) by Lemma 33. Applying ϕ to (2) gives (3) (xβ (t), xα (u2 )) = xα+2β (t2 u2 )xα+β (tu2 ). This is valid, since it can be obtained from (2) by taking inverses34 and replacing t by u2 , u by t. (c) The map ϕ in (b) is outer, for if we represent G as Sp4 and if t = 0, xα (t)−1 has rank 1 while xβ (t) − 1 has rank 2. (d) If in (b) |k| = 2, ϕ leads to an outer automorphism of S6 since, in fact, Sp4 (2) ∼ = S6 . To see this represent S6 as the Weyl group of type A5 . This fixes a bilinear form with matrix ⎡ ⎤ 2 −1 0 0 0 ⎢ −1 2 −1 0 0 ⎥ ⎢ ⎥ ⎢ 0 −1 2 −1 0 ⎥ ⎢ ⎥ ⎣ 0 0 −1 2 −1 ⎦ 0 0 0 −1 2 relative to a basis of simple roots. This is so because, to multiplication by a scalar,   2 the form is just xi xj (αi , αj ) = | xi αi | . Reduce mod 2. The line through α1 +α3 +α5 becomes invariant and the form becomes skew and nondegenerate on the quotient space. Hence we have a homomorphism ψ : S6 → Sp4 (2). It is easily seen that ker ψ  A6 so ker ψ = 1. Since |S6 | = 6! = 720 = 24 (22 − 1)(24 − 1) = |Sp4 (2)|, ψ is an isomorphism. ψ −1 may be described as follows. Sp4 (2) acts on the underlying projective space 3 P which contains 15 points. Given a point p there are 8 points not orthogonal to p. 34 Recall

that since char k = 2 we have xγ (u) = xγ (u)−1 for all γ ∈ Σ, u ∈ k. ()

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10. ISOMORPHISMS AND AUTOMORPHISMS

87

These split into two four point sets S1 , S2 such that each of {p} ∪ S1 and {p} ∪ S2 consists of mutually nonorthogonal points and these are the only five element sets containing p with this property. There are 15 · 2/5 = 6 such 5 element sets. Sp4 (2) acts faithfully by permutation on these 6 sets, so Sp4 (2) → S6 is defined. Under the outer automorphism the stabilizers of points and lines are interchanged. Each of the five point sets corresponds to a set of five mutually skew isotropic lines.35 Proof of Theorem 28. If p = 2 each α = 1. We must show that ϕ is defined on the xα (t) by the given equations preserves (A), (B), and (C). Here (A) and (C) follow at once. The nontrivial relations in (B) are: ⎧ ⎪ xα+β (±tu), if |α| = |β| and ∠(α, β) = 120◦ , ⎪ ⎪ ⎨ if α, β are short, orthogonal, and (xα (t), xβ (u)) = xα+β (±2tu), α + β ∈ Σ, ⎪ ⎪ ⎪ ⎩x 2 ◦ α+β (±tu)xα+2β (±tu ), if |α| > |β| and ∠(α, β) = 135 . (The last equation follows from Lemma 33. In the others the right hand side is of the form xα+β (Nα+β tu).) If p = 2 the second equation can be omitted and there are no ambiguities in sign. Because of the calculations in Example (b) above ϕ preserves these relations. Thus ϕ extends to a homomorphism. There remains only the case G2 , p = 3. The proof in that case depends on a sequence of lemmas. Lemma 56. Let G be a Chevalley group. Let α, β be distinct simple roots, n the order of wα wβ in W , so that wα wβ wα · · · = wβ wα wβ · · · (n factors on each side) in W . Then: (a) wα (1)wβ (1)wα (1) · · · = wβ (1)wα (1)wβ (1) · · · (n factors on each side) in G. (b) Both sides map Xα to −Xwα (where w = wα wβ · · · ). Proof. We may assume G is universal. For simplicity of notation we assume n = 3. Consider x = wα (1)wβ (1)wα (1)wβ (−1)wα (−1)wβ (−1). Let Gα = Xα , X−α . Then the product of the first five factors of x is in wα (1)wβ (1)Gα wβ (−1)wα (−1) = Gwα wβ α = Gβ and hence x ∈ Gβ . Similarly x ∈ Gα . By the uniqueness in Theorem 4 , x ∈ H. By the universality of G, x = 1. Let y = wα (1)wβ (1)wα (1). Then yXα = cX−β where c = ±1. Since [Xα , X−α ] = Hα is preserved by y, yX−α = cXβ (same c as above). Exponentiating and using wα (1) = xα (1)x−α (−1)xα (1) we obtain ywα (1)y −1 = w−β (c) = wβ (−c). By (a) ywα (1)y −1 = wβ (1), so c = −1, proving (b).  Lemma 57. If a, b are elements of an associative algebra over a field of characteristic 0, if both commute with [a, b] and if exp makes sense then exp(a + b) = (exp a)(exp b)(exp(−[a, b]/2)) Proof. Consider f (t) = (exp(−(a + b)t))(exp at)(exp bt)(exp(−[a, b]t2 /2)), a formal power series in t. Differentiating we get f  (t) = (−(a + b) + a − [b, a]t + b − [a, b]t)f (t) = 0. Hence f (t) = f (0) = 1.  35 Isotropic

lines in P3 are lines corresponding to isotropic planes in the 4-dimensional space.

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10. ISOMORPHISMS AND AUTOMORPHISMS

Now assume G is a Chevalley group of type G2 over a field of characteristic 0, and that the corresponding root system is as shown. 2α + 3β

α α+β

α + 2β α + 3β β

(1) Let y = wα (1)wβ (1) be an element of G corresponding to w = wα wβ (rotation through 60◦ (clockwise)). Then the Chevalley basis of L can be adjusted by sign changes so that yXγ = −Xwγ for all γ. Proof of (1). Let yXγ = cγ Xwγ , cγ = ±1. Then (∗) (∗∗)

cγ = c−γ ,

and

cγ cwγ cw2 γ = −1,

(by Lemma 56(b)). Adjust the signs of Xwα and Xw2 α so that cα = cwα = −1, and adjust the signs of X−wα and X−w2 α in the same way. It is clear from (∗) and (∗∗) that cγ = −1 for all γ in the w-orbit through α. Similarly we may make cγ = −1 for all γ in the w-orbit through β.  (2) (a) In (1) we have Nwγ,wδ = −Nγ,δ for all γ, δ. (b) We may arrange so that Nα,β = 1 and Nα+β,β = 2. It then follows that Nβ,α+2β = Nα+β,α+2β = 3 and Nα,α+3β = 1. Proof of (2). (a) follows from applying y to [Xγ , Xδ ] and using (1). In the proof of (b) we use (∗) if γ, δ are roots and {γ + iδ : −r ≤ i ≤ q} is the δ string through γ, then Nγ,δ = ±(r + 1), and Nγ,δ and Nγ+δ,−δ have the same sign (for their product is q(r + 1)). By changing the signs of all Xγ for γ in a w-orbit we can preserve the conclusion of (1) and arrange that Nα,β = 1, Nα+β,β = 2. By (a) and (∗) we have Nβ,α+2β = Nα+β,α+2β = 3N−α−β,2α+3β = −3Nβ,α = 3. Now [Xα , [Xα+2β , Xβ ]] = [Xα+2β , [Xα , Xβ ]], so that Nα+2β,β Nα,α+3β = Nα,β Nα+2β,α+β .  Hence Nα,α+3β = Nα,β = 1. (3) If (1) and (2) hold then: (a) (b) (c) (d) (e)

(xα (t), xβ (u)) = xα+β (tu)xα+3β (−tu3 )xα+2β (−tu2 )x2α+3β (t2 u3 ). (xα+β (t), xβ (u)) = xα+2β (2tu)xα+3β (−3tu2 )x2α+3β (3t2 u). (xα (t), xα+3β (u)) = x2α+3β (tu). (xα+2β (t), xβ (u)) = xα+3β (−3tu). (xα+β (t), xα+2β (u)) = x2α+3β (3tu).

Proof of (3). (a) By (2) xβ (u)Xα = (exp ad uXβ )Xα = Xα − uXα+β + u2 Xα+2β + u3 Xα+3β . Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

10. ISOMORPHISMS AND AUTOMORPHISMS

89

Multiplying by −t and exponentiating we get xβ (u)xα (−t)xβ (−u) = exp(−tXα − tu3 Xα+3β ) exp(tuXα+β − tu2 Xα+2β ) = xα (−t)xα+3β (−tu3 )x2α+3β (−t2 u3 /2)xα+β (tu)xα+2β (−tu2 )x2α+3β (3t2 u3 /2) by Lemma 57, which yields (a). The proof of (b) is similar. In (c) – (e) the term on the right hand side corresponds to the only root of the form iγ+jδ. The coefficient is Nγ,δ . We have taken the opportunity of working out all of the nontrivial relations of U explicitly. However, we will only use them in characteristic 3 when they simplify considerably.  (4) (a) There exists an automorphism θ of G such that if w is a rotation through 60◦ then θxγ (t) = xwγ (−t) for all γ ∈ Σ, t ∈ k. (b) If char k = 3, then there exists an endomorphism ϕ of G such that if r is the permutation of the roots given by rotation through 30◦ then ® xrγ (−t), if γ is long, ϕxα (t) = xrγ (t3 ), if γ is short. Proof of (4). (a) Take θ to be the inner automorphism by the element y of (1). (b) The relations (A) and (C) are clearly preserved. Now on the generators ϕ2 = θ ◦ ψ, where ψ : xα (t) → xα (t3 ), hence ϕ2 extends to an endomorphism of G. This implies that in verifying that the relations (B) are preserved by ϕ it suffices to show this for one pair of roots (γ, δ) with ∠(γ, δ) = each of the angles 30◦ , 60◦ , 90◦ , 120◦ , 150◦ . For if R(γ, δ) is the relation  (xγ (t), xδ (u)) = xiγ+jδ (cij ti uj ) if ∠(γ  , δ  ) = ∠(γ, δ), and if ϕ preserves R(γ, δ) then ϕ preserves R(γ  , δ  ). To show this it is enough to show that ϕ preserves R(rγ, rδ). If ϕ does not preserve R(rγ, rδ) then ϕ2 does not preserve R(γ, δ), a contradiction since ϕ2 = θ ◦ ψ extends to an endomorphism.36 If remains to verify R(γ, δ) for pairs of roots (γ, δ) with ∠(γ, δ) = 30◦ , 60◦ , 90◦ , ◦ 120 , 150◦ . For ∠(γ, δ) = 30◦ , 60◦ , 90◦ we take γ = α, δ = α + β, 2α + 3β, α + 2β respectively. Here we have commutativity both before and after applying ϕ (since (e) becomes trivial since char k = 3). For ∠(γ, δ) = 120◦ take (γ, δ) = (α, α + 3β). Then ϕ converts (c) to (xα+β (−t), xβ (−u)) = xα+2β (−tu) which is (b), a valid relation. For ∠(γ, δ) = 150◦ we take (γ, δ) = (α, β). We compare the constants Nγ,δ for the positive root system relative to (α, β) and the corresponding positive root system relative to (−α, α + β). Corresponding to Nα,β = 1 we have N−α,α+β = 1 and corresponding to Nα+β,β = 2 we have Nβ,α+β = −2. By changing the sign of 36 This is somewhat imprecise. If ϕ preserves the relation R(rγ, rδ) not everywhere but only on the image of ϕ (and preserves R(γ, δ)), then ϕ2 preserves R(γ, δ). However, this is easy to fix since in the verification of the fact that ϕ can be extended to an endomorphism one can replace the field k with its algebraic closure k. After that all arguments become true since ϕ becomes an epimorphism. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Xγ for all short roots γ we return to the original situation. Since wα maps the first system onto the second, ® xwα γ (−t), if γ is short, η : xγ (t) → xwα γ (t), if γ is long, extends to an automorphism of G, and so to prove that ϕ preserves R(γ, δ) it is sufficient to prove that η ◦ ϕ does, i.e., that (a) is preserved by ® xwα rγ (t), if γ is long, xγ (t) → xwα rγ (t3 ), if γ is short. (Note that wα r is the reflection in the line bisecting ∠(α, β).) I.e., that the following equations are consistent: (a) (a )

xα+β (tu) xα+3β (−tu3 ) xα+2β (−tu2 ) x2α+3β (t2 u3 ) OOOoo QQQmm woooOO' vmmmQQ( (xβ (t), xα (u3 )) = xα+3β (t3 u3 ) xα+β (−tu3 ) x2α+3β (−t3 u6 ) xα+2β (t2 u3 ) (xα (t), xβ (u)) =

Equation (a ) follows from (a) by replacing t by u3 , u by t, and taking inverses.37 This proves (4).  We now complete the proof of Theorem 28. The only remaining case of the first statement is G of type G2 , p = 3. If G = G∗ this follows from (4) above. In fact, whether G = G∗ or not is immaterial because (∗) a universal Chevalley group is determined by Σ and k independently of L or the Chevalley basis of L. (∗) follows from Theorem 29 below. Assume now that k is perfect. Then ϕ maps one set of generators one-to-one onto the other so that ϕ−1 exists on the generators. Since ϕ preserves (A), (B),  and (C) so does ϕ−1 . Hence ϕ−1 exists on G∗ , i.e., ϕ is an isomorphism. Remark. If k is not perfect, and ϕ : G → G, then ϕG is the subgroup of G in which Xα is parametrized by k if α is long, by kp if α is short. Here kp can be replaced by any field between kp and k to yield a rather weird simple group. Theorem 29. Let G and G be Chevalley groups constructed from (L, B = {Xβ , Hα : α ∈ Σ} , L, k)

and

(L , B  = {Xα , Hα : α ∈ Σ } , L , k)

respectively. Assume that there exists an isomorphism of Σ onto Σ taking α → α such that L maps onto L . Then there exists an isomorphism ϕ : G → G and signs α (α ∈ Σ) such that ϕxα (t) = xα (α t) for all α ∈ Σ, t ∈ k. Furthermore we may take α = +1 if α or −α is simple. Proof. By the uniqueness theorem for Lie algebras with a given root system there exists an isomorphism ψ : L → L such that ψXα = α Xα , ψHα = Hα with α in the base field for L (of characteristic 0) and α = 1 if α or −α is simple. (For this see, e.g., Jacobson [35].) By Theorem 1, Nα,β = ±(r + 1) = Nα ,β  . By induction on heights every α = ±1. Let ρ be a faithful representation of L used to construct G . Then ρ ◦ ψ is a representation of L which can be used to construct  G. Then xα (t) = xα (α (t)), so that ϕ = id meets our requirements. 37 We

use the fact that all terms in the right-hand side of (a) commute. ()

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Remarks. (a) Suppose k is infinite and we try to prove Theorem 28 with tp replaced by t. Then we must fail. For then the transpose of ϕ|H , mapping characters on H ∗ to those on H, maps Σ∗ onto Σ in the inversional manner of Lemma 55, hence can not be a homomorphism. This explains why long and short roots are treated differently. (b) If k is algebraically closed and we view G and G∗ as algebraic groups then ϕ is a homomorphism of algebraic groups and an isomorphism of abstract groups, but not an isomorphism of algebraic groups (for taking p-th roots (which is necessary for the inverse map) is not a rational operation). (c) For type G2 , char k = 3 (a similar result for C2 and F4 , char k = 2), in Lk there is an endomorphism dϕ such that ® if α is long, −Xrα , dϕ : Xα → 3Xrα = 0, if α is short. dϕ



Thus L −−→ Lshort −−→ 0 is exact, where Lshort is the 7-dimensional ideal spanned by all Xγ and Hγ for γ short. This leads to an alternate proof of the existence of ϕ. Corollary. (a) Let Σ be an indecomposable root system, σ an angle preserving permutation of the simple roots, σ = 1. If all roots are equal in length then σ extends to an automorphism of Σ. If not, and if p is defined as above, then σ must interchange long and short roots and σ extends to a permutation σ of all roots which also interchanges long and short roots and is such that the map α → σα if α is long, α → pσα if α is short is an isomorphism of root systems. The possibilities for σ are: (i) one root length:

An (n ≥ 2)

σ2 = 1

Dn (n ≥ 4)

σ2 = 1

D4

σ3 = 1

E6

σ2 = 1

(ii) two root lengths, σ 2 = 1 in all cases. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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C2

>

p=2

F4

>

p=2

G2

>

p=3

(b) Let k be a field and G a Chevalley group constructed from (Σ, k). Let σ be as in (a). If two root lengths occur assume k is perfect of characteristic p. If G is of type D2n , and char k = 2, assume σL = L. Then there exists an automorphism ϕ of G and signs α (α = 1 if α or −α is simple) such that ® xσα (α t), if α is long or all roots are of one length, ϕxα (t) = xσα (α tp ), if α is short. Proof. (a) is clear. (b) If G is universal the existence of ϕ follows from Theorems 28 and 29. If G is not universal let π : G → G be the universal covering. To show that ϕ can be dropped from G to G it is necessary to show that ϕ ker π ⊆ ker π. Now ker π ⊆ Z(G ) and unless G is of type D2n with char k = 2 the center of G is cyclic, so the result follows. Now suppose G is of type D2n and char k = 2. If C  = Z(G ), then C  is canonically isomorphic to Hom(L1 /L0 , k∗ ) = (L1 /L0 )∗ , giving a correspondence between subgroups C of C  and lattices L between L0 and L1 such that ϕC ⊆ C if and only if σL ⊆ L. Since ker π corresponds to L and σL = L, the result follows.  Remark. The preceding argument shows that for D2n in char k = 2, an automorphism of G fixing H and permuting the Xα ’s according to σ can exist only if σL = L. Remark. Automorphisms of G of this type as well as the identity are called graph automorphisms. Exercise. (a) Prove ϕ above is outer. (b) By imbedding A2 in G2 as the subgroup generated by all Xα such that α is long, show that its graph automorphisms can be realized by inner automorphisms of G2 . Similarly for D4 in F4 , Dn in Bn , and E6 in E7 . Lemma 58. Let G be a Chevalley group over k, fα ∈ k∗ for all simple α. Let f be extended to a homomorphism of L0 into k∗ . Then there exists a unique automorphism ϕ of G such that ϕxα (t) = xα (fα t) for all α ∈ Σ. Proof. Consider the relations (B), (xα (t), xβ (u)) = xiα+jβ (cij ti uj ). Applying ϕ we get the same thing with t replaced by fα t, u replaced by fβ u (for fiα+jβ = fαi fβj ). The relations (A) and (C) are clearly preserved. The uniqueness is clear.  Remark. Automorphisms of this type are called diagonal automorphisms. Exercise. Prove that every diagonal automorphism of G can be realized by conjugation of G in G(k) by an element in H(k). Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Example. Conjugate SLn by a diagonal element of GLn . If G is realized as a group of matrices and γ is an automorphism of k then the map γ : xα (t) → xα (tγ ) on generators extends to an automorphism of G. Such an automorphism is called a field automorphism. Theorem 30. Let G be a Chevalley group such that Σ is indecomposable and k is perfect. Then any automorphism of G can be expressed as the product of an inner, a diagonal, a graph and a field automorphism. Proof. Let σ be any automorphism of G. (1) The automorphism σ can be normalized by multiplication by an inner automorphism so that σU = U , σU − = U − . If this is done then σH = H and there exists a permutation ρ of the simple roots such that σXα = Xρα and σX−α = X−ρα for all simple α. Proof of (1). If k is finite, U is a p-Sylow subgroup (p = char k) by the corollary of Theorem 25, so by Sylow’s Theorem we can normalize σ by an inner automorphism so that σU = U . If k is infinite the proof of the corresponding statement is more difficult and will be given at the end of the proof (steps (5)– (12)). For now we assume σU = U . U − is conjugate to U , so σU − = uwU w−1 u−1 for some w ∈ W, u ∈ U . Since − U ∩ U = 1, uwU w−1 u−1 ∩ U = 1 and hence wU w−1 ∩ U = 1. Thus w = w0 so σU − = uU − u−1 . Normalizing σ by the inner automorphism corresponding to u−1 we get σU = U , σU − = U − . Now B = U H = normalizer of U , B − = U − H = normalizer of U − . Hence σ fixes B ∩ B − = H. Also σ permutes the (B, B) double cosets. Now B ∪ BwB (w = 1) is a group if and only if w = wα , α simple. Therefore σ permutes these groups. Now (B ∪ Bwα B) ∩ U − = X−α for B ∪ Bwα B = Bwα−1 ∪ Bwα Bwα−1 = Bwα−1 ∪ BX−α . Since B ∩ U − = 1, BX−α ∩ U − = X−α . We must show Bwα−1 ∩ U − is empty. Thus it suffices to show Bwα−1 w0 ∩ U − w0 = Bwα−1 w0 ∩ w0 U is empty. This holds by Theorem 4. Therefore the X−α ’s, α simple, are permuted by σ and similarly for the Xα ’s. The permutation in both cases is the same since Xα and X−β commute (α, β simple) if and only if α = β.  (2) The automorphism σ can be further normalized by a diagonal automorphism so that σxα (1) = xρα (1) for all simple α. It is then true that σx−α (1) = x−ρα (1) and σwα (1) = wρα (1). Further ρ preserves angles. In the proof of (2) we use: Lemma 59. Let α be a root, t ∈ k∗ , u ∈ k. Then xα (t)x−α (u)xα (t) = x−α (u)xα (t)x−α (u) if and only if u = −t

−1

, in which case both sides equal wα (t).

Proof. It suffices to verify this in SL2 , where it is immediate.



Proof of (2). We can achieve σxα (1) = xρα (1) for all simple roots α by a diagonal automorphism. By Lemma 59 with t = 1, σx−α (−1) = x−ρα (−1) and hence σwα (1) = wρα (1). Suppose α and β are simple roots. Then ∠(α, β) = π − π/n where n = (order of wα wβ in W ) = (order of wα (1)wβ (1) mod H) = (order of σ(wα (1)wβ (1)) mod H) = (order of wρα wρβ in W ). Hence ∠(α, β) = ∠(ρα, ρβ).  Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(3) σ can be further normalized by a graph automorphism so that ρ = 1. Proof of (3). By the corollary to Theorem 28 a graph automorphism exists corresponding to ρ provided that p = 2 if Σ is of type C2 or F4 , or p = 3 if Σ is of type G2 , or ρL = L if Σ is of type D2n and char k = 2, in the notation there. Suppose Σ is of type C2 or F4 and ρ = 1. Then there exist simple roots α and β, α long, β short, such that α + β and α + 2β are roots, ρα = β, ρβ = α, and wα (1)Xβ wα (−1) = Xα+β . Applying σ we get σXα+2β = Xα+β , σXα+β = Xα+2β . Since Xα and Xα+2β commute so do Xβ and Xα+β . Hence 0 = Nα+β,β = ±2. Hence char k = 2 so the required graph automorphism exists. Similarly it exists if Σ is of type G2 . Finally, if Σ is of type D2n , char k = 2, and ρ is extended in the obvious way, then ρL = L by the remark after Corollary (b) to Theorem 28, so that the graph automorphism exists by the corollary itself.  (4) σ can now be normalized by a field automorphism so that σ = 1 (i.e., if σ satisfies σU = U , σU − = U − , σxα (1) = xα (1) for all simple roots α then σ is a field automorphism. Proof of (4). Fix a simple root α and define f : k → k by σxα (t) = xα (f (t)). We will show that f is an automorphism of k. We have f additive, onto, f (1) = 1 and by Lemma 59 σwα (t) = wα (f (t)). Therefore σhα (t) = hα (f (t)). Since the kernel of the map t → hα (t) is contained in {±1} and hα (t) is multiplicative, f is multiplicative up to sign. Assume f (tu) = af (t)f (u) (where a = a(t, u) = ±1 and t, u = 0). We must show a = 1. Then af (t)f (u) + f (u) = f (tu) + f (u) = f ((t + 1)u) = bf (t + 1)f (u) (where b = b(t + 1, u) = ±1) = b(f (t) + 1)f (u) = bf (t)f (u) + bf (u). Hence (b − a)f (t) = 1 − b. Thus if a = b then a = b = 1. If a = b, then b = 1 so that a = 1 again. Hence f is an automorphism. Let β be another simple root connected to α in the Dynkin diagram (β may not exist). Let g be the automorphism of k corresponding to β. Then σ fixes Xα+β . Consider (xα (t), xβ (u)) = xα+β (±tu) · · · (+ or − is independent of t, u). Applying σ, first with u = 1 then with t = 1, u replaced by t we get σ(xα (t), xβ (1)) = (xα (f (t)), xβ (1)) = xα+β (±f (t)) · · · σ(xα (1), xβ (t)) = (xα (1), xβ (g(t)) = xα+β (±g(t)) · · · In either case the Xα+β term on the right is σxα+β (±t). Hence f (t) = g(t). Since Σ is indecomposable there is a single automorphism f of k so that σxα (t) = xα (f (t)) for all simple α. Applying the field automorphism f −1 to G we get the normalization f = 1, i.e., σ fixes every xα (t), wα (t) for α simple. These elements generate G so σ = 1.  We now assume k is infinite and consider the normalization σU = U of (1). (5) Assume that k is infinite, that A and M are its additive and multiplicative groups, that A0 and M0 are infinite subgroups such that A/A0 is finite and M/M0 is a torsion group. If M0 A0 ⊆ A0 then A0 = A. Proof of (5). Let F be the additive group generated by M0 . Then F is a field for it is closed under multiplication and addition and if f ∈ F , f = 0 then f −r ∈ F for some r > 0 so f −1 = f r−1 f −r ∈ F . Now for a ∈ A, a = 0, F a ∩ A0 is nontrivial since F a is infinite and A/A0 is finite. Thus f a ∈ A0 for some f = 0.  Hence a ∈ F A0 ⊆ A0 . Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

10. ISOMORPHISMS AND AUTOMORPHISMS

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(6) If B, U are as usual, k is infinite and B0 is a subgroup of finite index in B, then DB0 = U . Proof of (6). Fix α and identify subgroup of k) and  Xα with A (the additive  Xα ∩ B0 with A0 in (5). Set M0 = t2 : hα (t) ∈ α ∩ B0 . Now (∗)

hα (t)xα (u)hα (t)−1 = xα (t2 u)

so M0 A0 ⊆ A0 . M0 is infinite and M/M0 is torsion so by (5) Xα ∩ B0 = Xα , i.e., Xα ⊆ B0 . By (∗) DB0 ⊇ Xα . Thus DB0 ⊇ U . Since B0 /U is abelian DB0 ⊆ U .  (7) If A is a connected solvable algebraic group then DA is a connected unipotent group. This follows from: Theorem (Lie-Kolchin). Every connected solvable algebraic group A is reducible to superdiagonal form. Proof. We use induction on the dimension of the underlying space V and thus need only to find a common eigenvector and may assume V is irreducible. Let A1 = DA. By induction on the length of the derived series of A there exists v ∈ V , v = 0 such that x1 v = χ(x1 )v for all x1 ∈ A1 , χ a rational character on A1 . Let Vχ be the space of all such v. A normalizes A1 and hence permutes the Vχ , which are finite in number. Since A is connected this is the identity permutation. Since V is irreducible there is only one Vχ and it is all of V , i.e., A1 acts by scalars. Since A1 = DA each element of A1 has determinant 1 so there are only finitely many scalars. Since A1 is connected all scalars are 1, that is A1 acts trivially. Thus A/A1 is abelian and acts on V1 and hence has a common eigenvector.  An algebraic variety is complete if whenever it is imbedded densely in another variety it is the entire variety. (For a more exact definition see Mumford [45].) Examples. The affine line is not complete. It can be imbedded in the projective line. The following are complete: (a) All projective spaces. (b) All flag spaces. (c) B\G where G is a connected linear algebraic group and B is a maximal connected solvable subgroup. (See S´eminaire Chevalley [51], Exp. 5–10.) We now state, without proof, two results about connected algebraic groups acting on complete varieties. (8) Theorem (Borel) A connected solvable algebraic group acting on a complete variety fixes some point. This is an extension of the Lie–Kolchin theorem, which may be restated: every connected solvable algebraic group fixes some flag on the underlying space. We need a refinement of a special case of it. Theorem (Rosenlicht [47]). If A is a connected unipotent group acting on a complete variety V , if everything is defined over a perfect field k, and if V contains a point over k, then it contains one fixed by A. Notation. Let G be a Chevalley group over an infinite field k, k the algebraic closure of k, G, B constructed over k, and Gk the set of elements in G whose coordinates lie in k. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(9) The map Gk → (B\G)k is onto. Proof of (9). Assume Bx is defined over k, x = wu as in Theorem 4 . We can take w a product of wα (1)’s defined over k. Therefore Bu− is defined over k where u− = wuw−1 ∈ U − . Now U − is defined over k. Since u− = Bu− ∩ U − , u− is defined over k and hence x is defined over k.  (10) Gk = H k G. Proof of (10). See the proof of Theorem 7, Corollary 3.



(11) If A is a connected unipotent subgroup of G defined over k, it is Gconjugate to a subgroup of U . Proof of (11). We make A act on B\G by right multiplication. By (8) there exists Bx defined over k fixed by A. By (9) we can choose x ∈ Gk , and then by (10) x ∈ G. We have Bxa = Bx for all a ∈ A, i.e., xax−1 ∈ B for all a ∈ A, so that  xAx−1 ⊆ B. Since A is unipotent xAx−1 ⊆ U . (12) If k is infinite and perfect, the normalization σU = U of (1) can be obtained. Proof of (12). σB is solvable so σB (the smallest algebraic subgroup of G containing σB) is solvable. Hence (σB)0 , the connected component of the identity, is solvable and of finite index in σB. (σB)0 = σ(B0 ) for some B0 of finite index in B. Let A = DσB0 . By (7) A is connected, unipotent and defined over k. By (6) σU ⊆ A. By (11) there exists x ∈ G such that xAx−1 ⊆ U . Hence xσU x−1 ⊆ U , i.e., the normalization σU ⊆ U has been attained. Then U ⊆ σ −1 U . But U is maximal with respect to being nilpotent and containing no elements of the center  of G. (Check this.) Therefore σ −1 U = U so σU = U . This completes the proof of Theorem 30.



Corollary. If k is finite then Aut(G)/ Int(G) is solvable. Exercise. Let D be the group of diagonal automorphisms modulo those which are inner. Prove: (a) D ∼ = k∗ /k∗ ei where = Hom(L0 , k∗ )/ {Homomorphisms extendable to L1 } ∼ the ei are the elementary divisors of L1 /L0 . (b) If k is finite, D ∼ = C, the center of the corresponding universal group. (c) D = 1 if k is algebraically closed or if all ei = 1. Examples. (a) SLn . Every automorphism can be realized by a semilinear mapping of the underlying space composed with either the identity or the inverse transpose. I.e., every automorphism is induced by a collineation or a correlation of the underlying projective space.38 (b) Over R or Q every automorphism of E8 , F4 , or G2 is inner. (c) The triality automorphism 38 Definitions

can be found, e.g., in Artin [1, Chapter II, § 10]. ()

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97

exists for Spin8 and PSO8 , but not for SO8 if char k = 2. (d) Aside from triality every automorphism of SOn or PSOn (split form) is induced by a collineation of the underlying projective space P which fixes the basic quadric Q : xi xn+1−i = 0. If n is even, there exist two families of (n − 2)/2dimensional subspaces P entirely within Q (e.g., if n = 4 the two families of lines in the quadric surface x1 x4 + x2 x3 = 0). The graph automorphism occurs because these two families can be interchanged. Theorem 31. Let G, G be Chevalley groups relative to (Σ, k), (Σ , k ) Σ, Σ indecomposable, k, k perfect. Assume G and G are isomorphic. If finite, assume also char k = char k . Then k is isomorphic to k , and either isomorphic to Σ or else Σ, Σ are of type Bn , Cn (n ≥ 3) and char k = char k

with k is Σ is = 2.

Proof. As in (1) and (2) of the proof of Theorem 30 we can normalize σ so that σU = U  , σxα (1) = xρα (1), where now ρ is an angle preserving map of Σ onto Σ . Hence Σ ∼ = Σ or else Σ, Σ are Bn , Cn (n ≥ 3). As in (3) char k = char k = 2  in the second case. As in (4) k ∼ = k . Corollary. Over a field of char k = 2 the Chevalley groups of types Bn , Cn (n ≥ 3) are not isomorphic. Exercise. If rank Σ, rank Σ ≥ 2 then the assumption char k = char k can be dropped in Theorem 31. (Hint: if p = char k and rank Σ ≥ 2 then p makes the largest prime power contribution to |G|. If you get stuck see Artin [2].) (There are exceptions in case rank Σ, rank Σ ≥ 2 fails, e.g., SL2 (4) ∼ = PSL2 (5), SL3 (2) ∼ = PSL2 (7).)

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https://doi.org/10.1090//ulect/066/12

CHAPTER 11

Some Twisted Groups In this chapter we study the group Gσ of fixed points of a Chevalley group under an automorphism σ. We consider only the simplest case, in which σ fixes U, H, U − , N , hence acts on W = N/H and permutes the Xα ’s. Before launching into the general theory, we consider some examples: (a) G = SLn . If σ is a nontrivial graph automorphism it has the form σx = ax−1 a−1 (where x is the transpose of x and ⎡

1

⎢ a=⎣

2 ..

.

⎤ ⎥ ⎦

i = ±1). We see that σ fixes x if and only if xax = a. If a is skew, we get Gσ = Spn . If a is symmetric, we get Gσ = SOn (split form). The group SO2n in characteristic 2 does not arise here, but it can be recovered as a subgroup of SO2n+1 , namely the one “supported” by the long roots. Let t → t be an involutory automorphism of k having k0 as fixed field. If σ is now modified so that σx = ax−1 a−1 , then Gσ = SUn (split form). This last result holds even if k is a division ring provided t → t is an anti-automorphism. If V is the vector space over R generated by the roots and W is the Weyl group, then σ acts on V and W and has fixed point subspaces Vσ and Wσ . Wσ is a reflection group on Vσ with the corresponding “roots” being the projection on Vσ of the original roots. To see these facts, we write n = 2m + 1 or n = 2m and use the indices −m, −(m−1), . . . , m−1, m with the index 0 omitted in case n = 2m. If ωi is the weight on H defined by ωi : diag(a−m , . . . , am ) → ai , then the roots are ωi − ωj (i = j) and σωi = −ω−i . Vσ is thus spanned by {ωi = ω−i − ωi : i > 0}. Now w ∈ Wσ if and only if w commutes with σ, i.e., if and only if w(ωi − ωj ) = ωk − ωl implies w(ω−i − ω−j ) = ω−k − ω−l . We see that Wσ is the octahedral group acting on Vσ by all permutations and sign changs of the basis {ωi }. The projection of ωi − ωj (i, j = 0) on Vσ is 12 (ωi − ω−i − ωj + ω−j ) = 12 (±ωk ± ωl ) (k, l > 0) or = ±ωk (k > 0). If either i = 0 or j = 0, the projection of ωi − ωj is ± 12 ωk (k > 0). The projected system is of type Cm if n = 2m or BCm (a combination of Bm and Cm ) if n = 2m + 1. (b) G =  SO2n (split form, char k = 2). We take the group defined by the form f = 2 ni=1 xi x−i . We will take the graph automorphism to be σx = Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 99

100

11. SOME TWISTED GROUPS

a1 ax−1 a−1 a−1 1 , where





1

⎢ a=⎣

..

.

⎤ ⎥ ⎦,

1

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ a1 = ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣



1 ..

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

. 1 0 1 1 0 1 .. 

n

. 1

The corresponding form fixed by elements of Gσ is f = 2 i=2 xi x−i + x21 + x2−1 . Thus, Gσ fixes f  − f = (x1 − x−1 )2 and hence the hyperplane x1 − x−1 = 0. Gσ on this hyperplane is the group SO2n−1 . If we now combine σ with t → t as in (a), the form f  is replaced by n  (xi x−i + x−i xi ) + x1 x1 + x−1 x−1 f  = i=2

If we make the change of coordinates x1 replaced by x1 + tx−1 , x−1 replaced by x1 + tx−1 (t ∈ k, t = t), we see that f is replaced by n  xi x−i + 2(x21 + ax1 x−1 + bx2−1 ) 2 i=2 

and f is replaced by n  (xi x−i + x−i xi ) + (2x1 x1 + a(x1 x−1 + x−1 x1 ) + 2bx−1 x−1 ) i=2

where a = t + t and b = tt. Since these two forms have the same matrix, Gσ is SO2n over k0 re the new version of f . That is, Gσ is SO2n (k0 ) for a form of index n − 1 which has index n over k.39 Example. If n = 4, k = C, and k0 = R, Gσ is the Lorentz group (re f = x21 − x22 − x23 − x24 ). If we observe that D2 corresponds to A1 × A1 , we see that SL2 (C) and the 0-component of the Lorentz group are isomorphic over their centers. Thus, SL2 (C) is the universal covering group of the connected Lorentz group. Exercise. Work out D3 ∼ = A3 in the same way. For other examples see E. Cartan [11], No. 38, especially at the end. Aside from the specific facts worked out in the above examples we should note the following. In the single root length case, the fixed point set of a graph automorphism yields no new group, only an imbedding of one Chevalley group in another (e.g., Spn or SOn in SLn ). To get a new group (e.g., SUn ) we must use a field automorphism as well. Now to start our general development we will consider first the effect of twisting abstract reflection groups and root systems. Let V be a finite-dimensional real Euclidean vector space and let Σ be a finite set of nonzero elements of V satisfying 39 Here by the index of a form the author means the maximal dimension of an isotropic subspace, i.e., of a subspace where the form vanishes. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(1) α ∈ Σ implies cα ∈ / Σ if c > 0, c = 1. (2) wα Σ = Σ for all α ∈ Σ where wα is the reflection in the hyperplane orthogonal to α. (See Appendix I.) We pick an ordering on V and let P (respectively Π) be the positive (respectively simple) elements of Σ relative to that ordering. Suppose σ is an automorphism of V which permutes the positive multiples of the elements of each of Σ, P , and Π. (Automorphism means isometry here.) It is not required that σ fix Σ, although it will if all elements of Σ have the same length. Let ρ be the corresponding permutation of the roots. Note that σ is of finite order and normalizes W . Let Vσ and Wσ denote the fixed points in V and W respectively. If α is the average of the elements in the σ-oribt of α, then (β, α) = (β, α) for all β ∈ Vσ . Hence the projection of α on Vσ is α. Theorem 32. Let Σ, P, Π, σ etc. be as above. (a) The restriction of Wσ to Vσ is faithful. (b) Wσ |Vσ is a reflection group. (c) If Σσ denotes the projection of Σ on Vσ , then Σσ is the corresponding “root system”; i.e., {wα |Vσ : α ∈ Σσ } generates Wσ |Vσ and wα Σσ = Σσ . However, (1) may fail for Σσ . (d) If Πσ is the projection of Π on Vσ , then Πσ is the corresponding “simple system”; i.e., if multiples are cast out (in case (1) fails for Πσ ), then Πσ is linearly independent and the positive elements of Σσ are positive linear combinations of elements of Πσ . Proof. Denote the projection of V on Vσ by v → v. This commutes with σ and with all elements of Wσ . (1) If α ∈ Σ, then α = 0; indeed α > 0 implies α > 0. If α is positive, so are all the vectors in the σ-orbit of α. Thus, their average α is also positive. If α < 0, then α = −(−α) < 0. (2) Proof of (a). If w ∈ Wσ , w = 1, then wα < 0 for some root α > 0. Thus wα = wα < 0 and α > 0. So w|Vσ = 1. (3) Let π be a ρ-orbit of simple roots, let Wπ be the group generated by all wα (α ∈ π), let Pπ be the corresponding set of positive roots, and let wπ be the unique element of Wπ so that wπ Pπ = −Pπ . Then wπ ∈ Wσ and wπ |Vσ = wα |Vσ for any root α ∈ Pπ . To see this, first consider σwπ σ −1 ∈ Wπ . Since σwπ σ −1 Pπ = Pπ , then σwπ σ −1 = wπ by uniqueness, and wπ ∈ Wσ . Since ρ permutes the elements of π in a single orbit, the projections on Vσ of the elements of Pπ are all positive multiples of each other. It follows that if α is any element of Pπ , then wπ α = −α. If v ∈ Vσ with (v, α) = 0, then 0 = (v, β) = (v, β) for β ∈ π. Hence wπ v = v. Thus wπ |Vσ = wα |Vσ . (4) If ν is a ρ-orbit of roots and w ∈ Wσ then all elements of wν have the same sign. This follows from wσα = σwα for all α ∈ Σ, w ∈ Wσ . (5) {wπ : π a ρ-orbit of simple roots} generates Wσ . Let w ∈ Wσ with w = 1 and let α be a simple root such that wα < 0. Let π be the ρ-orbit containing α. By (4), wPπ < 0 (i.e., wβ < 0 for all β ∈ Pπ ). Now wwπ Pπ > 0 and wπ permutes the elements of P − Pπ . Hence N (wwπ ) = N (w) − N (wπ ) (see Appendix II.17). Using induction on N (w), we may thus show that w is a product of wπ ’s. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(6) If w0 is the element of W such that w0 P = −P then w0 ∈ Wσ . This follows from σw0 σ −1 P = −P and the uniqueness of w0 . (7) {wPπ : w ∈ Wσ , π a ρ-orbit of simple roots} is a partition of Σ. If the wPπ ’s are called parts, then α, β belong to the same part if and only if α = cβ for some c > 0. To prove (7), we consider α ∈ Σ, α > 0. Now w0 α < 0 and w0 = w1 w2 · · · wr where each wi = wπ for some ρ-orbit of simple roots π (by (5) and (6)). Choose i so that wi+1 · · · wr α > 0 and wi wi+1 · · · wr α < 0. If wi = wπ , then wi+1 · · · wr α ∈ Pπ ; i.e., α is in some part. Similarly, if α < 0, α is in some part. Now assume α, β belong to the same part, say to wPπ . We may assume α, β ∈ Pπ . Then α and β are positive multiples of each other, as has been noted in (3). Conversely, assume that α is a root with support40 in a ρ-orbit π ∈ Π. If α = cβ, c > 0, then β has its support in π and hence so does β since σ maps simple roots not in π to positive multiples of simple roots not in π. We see that β ∈ Pπ , and that any part containing α also contains β. The parts are just the sets of β such that β = cα, c > 0 and hence form a partition. (8) {wα : w ∈ Wσ , α has support in a ρ-orbit of simple roots} = Σσ . (9) Parts (b) and (c) follow from (3), (5), and (8). (10) Proof of (d). We select one root α from each ρ-orbit and form {α}. This set, consisting of elements whose supports in Π are disjoint, is independent since Π is. If α > 0 then it is a positive linear combination of the elements of Π. Hence α  is a positive linear combination of the elements of Πσ . Remark. To achieve condition (1) for a root system, we can stick to the set of shortest projections of the various directions. Examples. 41 (a) For σ of order 2, W of type A2n−1 , we get Wσ of type Cn . For W of type A2n , we get Wσ of type BCn . (b) For σ or order 2, W of type Dn , we get Wσ of type Bn−1 . (c) For σ of order 3, W of type D4 , we get Wσ of type G2 . To see this let α, β, γ, δ be the simple roots with δ connected with α, β and γ. Then α = 13 (α + β + γ), δ = δ, and α, δ = −1, δ, α = −3, giving Wσ of G2 . Schematically: −→

=

>

(d) For σ of order 2, W of type E6 , we get Wσ of type F4 . =

−→

> support  of a vector v ∈ V is the set of simple roots α that have nonzero coefficients in nα · α. () the expansion, v = 41 These examples show that each crystallographic root system can be obtained as a twist of a crystallographic system with all roots of the same length. () 40 The

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(e) For σ of order 2, W of type C2 , we get Wσ of type A1 . (f) For σ of order 2, W of type G2 , we get Wσ of type A1 . (g) For σ of order 2, W of type F4 , we get Wσ of type D16 (the dihedral group of order 16). To see this we let

>

γ δ √ √ √ be the Dynkin diagram of F4 , and σα = 2δ, σβ = 2γ. Since α = 12 (α + 2δ), √ √ β = 12 (β + 2γ), we have β, α = −1, α, β = −(2 + 2). This corresponds to an angle of 7π/8 between α and β. Hence Wσ is of type D16 . Alternately, we note that wα wβ makes six positive roots negative and that there are 24 positive roots in all, so that w0 = (wα wβ )4 . Hence, wα2 = wβ2 = (wα wβ )8 = 1 and Wσ is of type D16 . Note that this is the only case of those we have considered in which Wσ fails to be crystallographic (See Appendix V). α

β

In (e), (f), (g) we are assuming that multiples have been cast out. The partition Σ in (7) above can be used to define an equivalence relation R on Σ by α ≡ β if and only if α is a positive multiple of β where α is the projection of α on Vσ . Letting Σ/R denote the collection of equivalence classes we have the following: Corollary. If Σ is crystallographic and indecomposable, then an element of Σ/R is the positive system of roots of a system of one of the following types: (a) An1 , n = 1, 2 or 3. (b) A2 (this occurs only if Σ is of type A2n ). (c) C2 (this occurs if Σ is of type C2 or F4 ). (d) G2 . Now let G be a Chevalley group42 over a field k of characteristic p. Let σ be an automorphism of G which is the product of a graph automorphism and a field automorphism θ of k and such that ρ is the corresponding permutation of the roots then (1) if ρ preserves lengths, then order θ = order ρ. (2) if ρ doesn’t preserve lengths, then pθ 2 = 1 (where p is the map x → xp ). (Condition (1) focuses our attention on the only interesting case. Observe that ρ = id, θ = id is allowed. Condition (2) could be replaced by θ 2 = p thereby extending the development to follow, suitably modified, to imperfect fields k.) We know that p = 2 if G is of type C2 or F4 and p = 3 if G is of type G2 . Recall also that ® xρα (α tθ ), if |α| ≥ |ρα|, σxα (t) = xρα (α tpθ ), if |α| < |ρα|, where α = ±1 and α = 1 if ±α is simple. (See the proof of Theorem 29.) Now σ preserves U, H, B, U − , and N , and hence N/H ∼ = W . The action thus induced on W is concordant with the permutation ρ of the roots. Since ρ 42 When

studying the group Gσ the general case is reduced to the case where the group G (i.e., the corresponding root system Σ) is indecomposable. Below we assume this to hold without mentioning it explicitly. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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preserves angles, it agrees up to positive multiples with an isometry on the real space generated by the roots. Thus the results of Theorem 32 may be applied. Also we observe that if n is the order or ρ, then n = 1, 2 or 3, so that the length of each ρ-orbit is 1 or n. Lemma 60. α = 1 over each ρ-orbit of length n. Proof. Since σ n acts on each Xα (±α simple) as a field automorphism, it does so on all of G, whence the lemma.  Lemma 61. If a ∈ Σ/R, then Xa,σ = 1.43 Proof. Choose α ∈ a so that no β ∈ a can be added to it to yield another root. If the orbit α has length 1, set x = xα (1) if α = 1, x = xα (t) with t ∈ k, t = 0 and t + tθ = 0 if α = 1. Then x ∈ Xa,σ . If the length is n, we set y = xα (1),  then x = y · σy · σ 2 y · · · over the orbit, and use Lemma 60. Theorem 33. Let G, σ, etc. be as above. (a) For each w ∈ Wσ , the group Uw = U ∩ w−1 U − w is fixed by σ. (b) For each w ∈ Wσ , there exists nw ∈ Nσ , indeed nw ∈ Uσ , Uσ− , so that nw H = w.  (c) If nw (w ∈ Wσ ) is as in (b), then Gσ = w∈Wσ Bσ nw Uw,σ with uniqueness of expression on the right. Proof. (a) This is clear since U and w−1 U − w are fixed by σ. (b) We may assume that w = wπ for some ρ-orbit of simple roots π. By Lemma 61, choose x ∈ X−a,σ x = 1, where a ∈ Σ/R corresponds to π. Using Theorem 4 we may write x = unw v for some w ∈ W where u ∈ U , v ∈ Uw , and nw H = w. Now x = σx = σu · σnw σv and by Theorem 4 and the uniqueness in Theorem 4 we have σw = w, σnw = nw , σu = u, and σv = v. Thus nw ∈ Uσ , Uσ− . Since w = 1, w ∈ Wσ , and w ∈ Wπ , we have wα < 0 for some α ∈ π, wπ < 0, and w = wπ . (c) Let x ∈ Gσ , say x ∈ BwB. Since σ(BwB) = BσwB we have w ∈ Wσ . Choose nw as in (b) and write x = bnw v with b ∈ B and v ∈ Uw . Applying σ we  get b ∈ Bσ and v ∈ Uw,σ . Uniqueness follows from Theorem 4 . Corollary. The conclusions of Theorem 33 are still valid if Gσ and Bσ are replaced by Gσ = Uσ , Uσ−  and Bσ = Gσ ∩ Bσ . Also since Bσ = Uσ Hσ , we can replace Hσ by Hσ = Gσ ∩ Hσ . Lemma 62. Let a generically denote a class in Σ/R. Let S be a union of classes in Σ/R which is closed under addition and such that if a ⊆ S then −a  S. Then XS,σ = a⊆S Xa,σ with the product taken in any fixed order and there is = uniqueness of expression on the right. In particular, U σ a>0 Xa,σ and Uw,σ = a>0,wa≤0 Xa,σ for all w ∈ Wσ . Proof. We arrange the positive roots in a manner consistent with the order of the a’s; i.e., those roots in the first a are first, etc. Now XS = α>0 Xα in the order just described and with uniqueness on the right by Lemma 17. Hence XS = a>0 Xa in the given order and again with uniqueness of expression on the 43 Recall that X is defined in Chapter 3. Here and later A (where A ⊂ G) is the intersection α σ of A and Gσ . () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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right. The lemma follows by considering the fixed points of σ on both sides of the last equation.  Corollary. If a, b are classes in Σ/R with a = ±b, then (Xa , Xb ) ⊆ Xc , where the roots on the right are in the closed subsystem generated by a and b, those of a and b excluded. The condition on c can be stated alternately, in terms of Σσ , that c is in the interior of the (plane) convex cone generated by a and b. Remark. The exact relations in the above corollary can be quite complicated but generally resemble those in the Chevalley group whose Weyl group is Wσ . For example, if G is of type A3 and σ is of order 2, say α

β

γ

a = {β}, b = {α, γ}, c = {α + β, β + γ}, d = {α + β + γ}, and if we set xa (t) = xβ (t) (t ∈ kθ ), xb (u) = xα (u)xγ (uθ ) (u ∈ k), and similarly for c and d we get (xa (t), xb (u)) = xc (±tu)xd (±tuuθ ). In C2 , the corresponding relation is (xa (t), xb (u)) = xa+b (±tu)xa+2b (±tu2 ). If G is of type X and σ is of order n, we say Gσ is of type nX . e.g., the group considered in the above remark is of type 2 A3 . The group of type 2 C2 is called the Suzuki group and the groups of type 2 G2 and 2 F4 are called Ree groups. We write G ∼ X and Gσ ∼ n X. Lemma 63. Let a be a class in Σ/R, then Xa,σ has the following structure: (a) If a ∼ A1 , then Xa,σ = {xa (t) : t ∈ kθ }. (b) If a ∼ An1 , then Xa,σ = {x · σx · · · : x = xα (t), α ∈ a, t ∈ k}. (c) If a ∼ A2 , a = {α, β, α + β}, then θ 2 = 1 and   Xa,σ = xα (t)xβ (tθ )xα+β (u) : ttθ + u + uθ = 0 . If (t, u) denotes the given element, then (t, u)(t , u ) = (t + t , u + u − tθ t ). (d) If a ∼ C2 , a = {α, β, α + β, α + 2β}, then 2θ 2 = 1 and   Xa,σ = xα (t)xβ (tθ )xα+2β (u)xα+β (t1+θ + uθ ) : t, u ∈ k . If (t, u) denotes the given element, (t, u)(t , u ) = (t + t , u + u + t2θ t ). (e) If a ∼ G2 , a = {α, β, α + β, α + 2β, α + 3β, 2α + 3β}, then 3θ 2 = 1 and Xa,σ = {xα (t)xβ (tθ )xα+3β (u)xα+β (uθ − t1+θ )x2α+3β (v)xα+2β (v θ − t1+2θ ) : t, u, v ∈ k}. If (t, u, v) denotes the given element then (t, u, v)(t , u , v  ) = (t + t , u + u + t t3θ , v + v  − t u + t2 t3θ ). Note that in (a) and (b), Xa,σ is a one parameter group for the fields kθ and k respectively. Proof. (a) and (b) are easy and we omit their proofs. For (c) normalize the parametrization of Xα+β so that Nα,β = 1. Then σxα (t) = xβ (tθ ), σxβ (t) = xα (tθ ), and σxα+β (u) = xα+β (−uθ ). Write x ∈ Xa,σ as x = xα (t)xβ (v)xα+β (u) and compare the coefficients on both sides of x = σx to get (c). The proof of (d) is similar to that of (c). For (e), first normalize the signs as in Theorem 28, and then complete the proof as in (c) and (d).  Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Exercise. Complete the details of the above proof. Remark. The role of the group SL2 in the untwisted case is taken by the groups SL2 (kθ ), SL2 (k), SU3 (k, θ) (split form), the Suzuki group, and Ree group of type G2 . Exercise. Determine the structure of Hσ in the case G is universal. Lemma 64. If G is universal, then Gσ is generated by Uσ and Uσ− except perhaps for the case Gσ ∼ 2 G2 with k infinite. Proof. Let Gσ = Uσ , Uσ−  and let Hσ = Hσ ∩ Gσ . By the corollary to Theorem 33, it suffices to show Hσ ⊆ Gσ ; i.e., (∗) Hσ = Hσ . Since G is universal, H is a direct product of {α : α simple} (see the corollary to Lemma 28). These groups are permuted by σ exactly as the roots are. Hence it is enough to prove (∗) when there is a single orbit; i.e., when Gσ is one of the types SL2 , 2 A2 , 2 C2 , or 2 G2 . For SL2 , this is clear. (1) For x ∈ Uσ − {1}, write x = u1 nu2 with ui ∈ Uσ− (i = 1, 2) and n= n(x) ∈ N ∩ Gσ . Then Hσ is generated by n(x)n(x0 )−1 : x0 a fixed choice of x . To see this let Hσ be the group so generated. Consider Gσ = Uσ− Hσ ∪ Uσ− Hσ n(x0 )Uσ− . This set is closed under right multiplication by Uσ− . It is also closed under right mul−1 −1 n(x0 ) tiplication by n(x0 )−1 . This follows from n(x0 )−1 = n(x−1 0 ) = n(x0 )n(x0 ) − −1  −1 = Uσ ⊆ Gσ since x = u1 (n(x)n(x0 ) )n(x0 )u2 for x ∈ and n(x0 )Uσ n(x0 ) Uσ − {1}. We see that Gσ = Gσ , whence Hσ = Hσ . (2) If α and β are the simple roots of A2 , C2 , or G2 labeled as in Lemma 63(c), (d), or (e) respectively, then Hσ is isomorphic to k∗ via the map ϕ : t → hα (t)hβ (tθ ). (3) Let λ be the weight such that λ, α = 1, λ, β = 0, let R be a representation of Lk (obtained from one of L by shifting the coefficients to k) having λ as highest weight and let v + be the corresponding weight vector. Let μ be the lowest weight of R and let v − be the corresponding weight vector. For x ∈ Uσ − {1}, write xv − = f (x)v + + terms for lower weights. Then f (x) = 0 and Hσ is isomorphic under ϕ−1 in (2) to the subgroup m of k∗ generated by all f (x)f (x0 )−1 . To prove (3), let x ∈ Uσ − {1} and write x = u1 n(x)u2 as in (1). We see xv − = n(x)v + + terms for lower weights, so n(x)v − = f (x)v + and n(x)n(x0 )−1 v + = f (x)f (x0 )−1 v + . If n(x)n(x0 )−1 = hα (t)hβ (tθ ), then by the choice of λ, f (x)f (x0 )−1 = t (see Lemma 19(c)). (3) then follows from (1). (4) The case Gσ ∼ 2 A2 . Here f (x) = −uθ and m = k∗ . To see this, we note that the representation R of (3) in this case is R : Lk → sl3 (k) and if x = xα (t)xβ (tθ )xα+β (u) then ⎡ ⎤ 1 t u + ttθ ⎦. tθ x → ⎣ 0 1 0 0 1 Thus, f (x) = u + ttθ = −uθ by Lemma 63(c). Thus, m is the group generated by ratios of elements (−uθ ) of k∗ whose traces are norms (ttθ ). Let u ∈ k∗ . If uθ = u, set u1 = (u − uθ )−1 , and if uθ = u, choose u1 ∈ k∗ so that uθ1 = −u1 . Then uu1 and u1 are values of f (their traces are 0 or 1), so that u ∈ m and m = k∗ . (5) The case Gσ ∼ 2 C2 . Here f (x) = t2+2θ + u2θ + tu and m = k∗ . To see this, first note that since char k = 2, there is an ideal in Lk “supported” by short roots. The representation R can be taken as Lk acting on this ideal and v + = Xα+β while Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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v − = X−α−β . Letting x = xα (t)xβ (tθ )xα+2β (u)xα+β (uθ + t1+θ ) we can determine f (x). By taking t = 0 in the expression for f (x) and writing v = (v θ )2θ , we see that m = k∗ . (6) The case Gσ ∼ 2 G2 . Here f (x) = t4+6θ − u1+3θ − v 2 + t3+3θ u + t1+3θ u3θ + tv − tuv. The group m is generated by all values of f for which (t, u, v) = (0, 0, 0), and it contains k∗ 2 and −1; hence m = k∗ , if k is finite. Here the representation R can be taken to be the adjoint representation of Lk , v + = X2α+3β , and v − = X−2α−3β . Letting x be as in Lemma 63(e), and working modulo the ideal in Lk “supported” by the short roots, we can compute f (x). Setting t = u = 0, we see that −v 2 ∈ m, hence −1 ∈ m and k∗ 2 ⊆ m. If k is finite m = k∗ follows from 3θ

(∗)

−1∈ / k∗ 2 . 2

To show (∗), suppose t2 = −1 with t ∈ k. Then t2θ = −1, so tθ = ±t and tθ = t. 2 Since 3θ 2 = 1, we see t = (tθ )3 = t3 . But t3 = t2 t = −t, so t = 0, a contradiction. This proves the lemma.  Corollary. If G is universal, then Gσ = Gσ and Hσ = Hσ except possibly for G2 with k infinite in which case Gσ /Gσ = Hσ /Hσ ∼ = k∗ /m with m as in (6) above. 2

Remarks. (a) It is not known whether m = k∗ always if Gσ ∼ 2 G2 . One can make the changes in variables v → v + tu and then u → u − t1+3θ to convert the form f in (6) to t4+6θ − u1+3θ − v 2 + t2 u2 + tv 3θ . Both before and after this simplification the form satisfies the condition of homogeneity: f (t, u, v) = t4+6θ f (1, u/t1+3θ , v/t2+3θ ) if

t = 0.

(b) A corollary of (3) above, is that the forms in (5) and (6) are definite, i.e., f = 0 implies t = u (= v) = 0. A direct proof in case f is as in (5) can be made as follows: Suppose 0 = f (t, u) = t2+2θ + u2θ + tu with one of t, u nonzero. If t = 0, then u = 0, so we have t = 0. We see f (t, u) = t2+2θ f (1, u/t2θ+1 ) using 2θ 2 = 1. Hence we may assume t = 1. Thus, 1 + u2θ + u = 0 or (by applying θ) uθ = 1 + u. 2 2 Hence uθ = 1 + uθ = u and u = u2θ = u2 . Thus, u = 0 or 1, a contradiction. A direct proof in case f is as in (6) appears to be quite complicated. (c) The form in (5) leads to a geometric interpretation of 2 C2 . Form the graph v = t2+2θ + u2θ + tu in k3 of the form f (x). Imbed k3 in P3 (k), projective 3-space over k, by adding a plane at ∞, and adjoin the point at ∞ in the direction (0, 0, 1) to the graph to obtain a subset Q of P3 (k). Q is then an ovoid in P3 (k); i.e., (1) No line meets Q in more than two points. (2) The line through any point of Q not meeting Q again always lie in a plane. The group 2 C2 is then realized as the group of projective transformations of P3 (k) fixing Q. For further details as well as a corresponding interpretation of 2 G2 see Tits [62]. For an exhaustive treatment of 2 C2 , especially in the finite case, see L¨ uneburg [41]. Theorem 34. Let G and σ be as above with G universal. Excluding the cases: (a) 2 A2 (4), (b) 2 B2 (2), (c) 2 G2 (3), (d) 2 F4 (2), we have that Gσ is simple over its center. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Sketch of proof. Using a calculus of double cosets re Bσ , which can be developed exactly as for Chevalley groups with Wσ in place of W and Σ/R (or Σσ (see Theorem 32)) in place of Σ, and Theorem 33, the proof can be reduced exactly as for the Chevalley groups to the proof of: Gσ = DGσ . If k has “enough” elements, so does Hσ by the corollary to Lemma 64 and the action of Hσ on Xa,σ can be used to show Xa,σ ⊆ DGσ . This takes care of nearly everything. If k has “few” elements then the commutator relations within the Xa ’s and among them can be used. This leads to a number of special calculations. The details are omitted.  Remark. The groups in (a) and (b) above are solvable. The group in (c) contains a normal subgroup of index 3 isomorphic to A1 (8). The group in (d) contains a “new” simple normal subgroup of index 2. (See Tits [63].) Exercise. Z(Gσ ) = (Z(G))σ . We are now going to determine the orders of the finite Chevalley groups of twisted type. Let k be a finite field of characteristic p. Let a be minimal such that a θ = pa (i.e., such that tθ = tp for all t ∈ k). Then |k| = p2a for 2 An , 2 Dn , 2 E6 ; |k| = p3a for 3 D4 ; and |k| = p2a+1 for 2 C2 , 2 F4 , 2 G2 . We can write σxα (t) = xρα (α tq(α) ) where q(α) is some power of p less than |k|. If q is the geometric average of q(α) over each ρ-orbit then q = pa except when Gσ is of type 2 C2 , 2 F4 , or 2 G2 in which case q = p(a+1)/2 . Let V be the real Euclidean space generated by the roots and let σ0 be the automorphism of V permuting the rays through the roots as ρ permutes the roots. Since σ0 normalizes W , we see that σ0 acts on the space I of polynomials invariant under W . Since σ0 also acts on the subspace of I of homogeneous elements of a given positive degree, we may choose the basic invariants Ij , j = 1, . . . , l of Theorem 27 such that σ0 Ij = j Ij for some j ∈ C (here we have extended the base field R to C). As before, we let dj be the degree of Ij , and these are uniquely determined. Since σ0 acts on V , we also have the set {0j : j = 1, . . . , l} of eigenvalues of σ0 on V . We recall also that N denotes the number of positive roots in Σ. Theorem 35. Let σ, q, N, j , and dj be as above, and assume G is universal. We have (a) |Gσ | = q N j (q dj − j ). (b) The order of the corresponding simple group is obtaining by dividing |Gσ | by |Cσ | where C is the center of G. Lemma 65. Let σ, H, U , etc. be as above. (a) |Uσ | = q N , |Uw,σ | = q N (w) . (b) |Hσ | = j (q − 0j ).  (c) |Gσ | = q N j (q − 0j ) w∈Wσ q N (w) . where N (w) is the number of positive roots in Σ made negative by w. Proof. (a) It suffices to show that |Xa,σ | = q |a| for a ∈ Σ/R by Lemma 62. This is so by Lemma 63. hρα (tq(α) ), the contribu(b) Let π be a ρ-orbit of simple roots. Since σhα (t) = tion to |Hσ | made by elements of Hσ “supported” by π is ( α∈π q(α)) − 1 = q m − 1 if m = |π|. Since the 0j ’s corresponding to π are the roots of the polynomial X m − 1, (b) follows. (c) This follows from (a), (b), and Theorem 33.  Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

11. SOME TWISTED GROUPS

109

Corollary. Uσ is a p-Sylow subgroup. Lemma 66. We have the following formal identity in t:   tN (w) = (1 − j tdj )/(1 − 0j t). w∈Wσ

j

Proof. We modify the proof of Theorem 26 as follows: (a) σ there is replaced by σ0 here. (b) Σ there is replaced by Σ0 here, where Σ0 is the set of unit vectors in V which lie in the same directions of the roots. (c) Only those subsets π of Π fixed by σ0 are considered. (d) (−1)π is now defined to be (−1)k where k is the number of σ0 orbits in π.  (e) W (t) is now defined to be w∈Wσ tN (w) . With these modifications the proof proceeds exactly as before through step (5). Steps (6)–(8) become: (6 ) For π ⊆ Π, w ∈ W , let Nπ be the number of cells in K congruent to Dπ under W and fixed by wσ0 . Then (−1)π Nπ = det w. (Hint: If V  = Vwσ0 and K  is the complex on V  cut by K, then the cells of K  are the intersections with V  of the cells of K fixed by wσ0 .) (7 ) Let χ be the character on W, σ0  and χπ the restriction of χ to Wπ , σ0   induced up to W, σ0 . Then (−1)π χπ (wσ0 ) = χ(wσ0 )(det w) (w ∈ W ).

) be the space of skew invariants under (8 ) Let M be a W, σ0  module, let I(M W , and let Iπ (M ) be the space of invariants under Wπ . Then 

)). (−1)π tr(σ0 , Iπ (M )) = tr(σ0 , I(M

The remainder of the proof proceeds as before.



Lemma 67. The j ’s form a permutation of the 0j ’s. Proof. Set t = 1 in Lemma 66. Then (∗) 1 has the same multiplicity among the j ’s as among the 0j ’s. This is so since otherwise the right side of the expression would have either a root or a pole at t = 1. Assume σ0 = 1, then either σ02 = 1 and all ’s not 1 are −1 or else σ03 = 1 and all ’s not 1 are cube roots of 1 coming in conjugate complex pairs, since σ0 is real. Thus in all cases (∗) implies the lemma.  Proof of Theorem 35. (a) follows from Lemma 65, 66, 67. Now let C  be the center of Gσ . Clearly C  ⊇ Cσ . Using the corollary to Theorem 33 and an argument similar to that in the proof of Corollary 1(b) to Theorem 4 , we see C  ⊆ Hσ ⊆ H. Since H acts “diagonally”, we have C  ⊆ C, hence C  = Cσ , proving (b).  Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Corollary. The values of |Gσ | and |Cσ | = |Hom(L0 /L1 , k∗ )| are as follows: Gσ Chevalley group (σ = 1)

j ’s = 1 None

An (n ≥ 2)

2

|Gσ | dj (q − 1)

|Cσ | |Hom(L0 /L1 , k∗ )|

−1 if dj is odd

Replace q dj − 1 by q dj − (−1)dj in (∗)

Same change; i.e., (n + 1, q + 1)

E6

Same as 2 An

Same as 2 An

(3, q + 1)

2

Dn

−1 for dj = n

Replace one q n − 1 by q n + 1 in (∗)

(4, q n + 1)

3

D4

ω, ω 2 for dj = 4, 4

q 12 (q 2 − 1)(q 6 − 1) · (q 8 + q 4 + 1)

1

2

(∗) q N

2

C2

−1 for dj = 4

q 4 (q 2 − 1)(q 4 + 1)

1

2

G2

−1 for dj = 6

q 6 (q 2 − 1)(q 6 + 1)

1

−1 for dj = 6, 12 q 24 (q 2 − 1)(q 6 + 1)(q 8 − 1) · (q 12 + 1) Here ω denotes a primitive cube root of 1. 2

F4

1

Proof (except for |Cσ |). We consider the cases: 2 An . We first note (∗) −1 ∈ W σ0 . To prove (∗) we use the standard coordinates {ωi : 1 ≤ i ≤ n + 1} for An . Then σ0 is given by ωi → −ωn+2−i . Since W acts via all permutations of {ωi }, we see −1 ∈ W σ0 . Alternatively, since W is transitive on the simple systems (Appendix II.24), there exists w0 ∈ W such that w0 (−Π) = Π. Hence w0 (−1) = 1 or σ0 ; i.e., −1 ∈ W or −1 ∈ W σ0 . Since there are invariants of odd / W . By (∗) σ0 fixes the invariants of even degree and degree di = 2, 3, . . .), −1 ∈ changes the signs of those of odd degree. 2 E6 , 2 D2n+1 . The second argument to establish (∗) in the case 2 An may be used here, and the same conclusion holds. 2 Dn (n even or odd). Relative to the standard coordinates {vi : 1 ≤ i ≤ n}, the 2 basic invariants are the first n − 1 elementary symmetric polynomials in vi together with vi , and W act via all permutations and even number of sign changes. Here σ0 can be taken to be the map vi → vi (1 ≤ i ≤ n − 1), vn → −vn . Hence, only the last invariant changes sign under σ0 . 3 D4 . The degrees of the invariants are 2, 4, 6, and 4. By Lemma 67, the j ’s are 1, 1, ω, ω 2 . Since σ0 is real, ω and ω 2 must occur in the same dimension. Thus, we replace (q 4 − 1)2 in the usual formula by (q 4 − ω)(q 4 − ω 2 ) = q 8 + q 4 + 1. 2 C2 , 2 G2 . In both cases the j ’s are 1, −1 by Lemma 67. Since W, σ0  is a finite group, it fixes some nonzero quadratic form, so that j = 1 for dj = 2. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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F4 . The degrees of the invariants are 2, 6, 8, 12 and the j ’s are 1, 1, −1, −1. As before there is a quadratic invariant fixed by σ0 . Consider I = α long root α8 + √  8 44 We claim that I is an invariant of degree 8 fixed by σ0 and β short root ( 2β) . there is a quadratic invariant fixed by σ0 which does not divide I. The first part is clear since W and σ0 preserve lengths and permute the rays through the roots. To see the second part, choose coordinates {vi : i = 1, 2, 3, 4} so that the long roots (respectively, the short roots) are the vectors obtained from 2v1 , v1 + v2 + v3 + v4 (respectively, v1 +v2 ) by all permutations and sign changes. The quadratic invariant is v12 + v22 + v32 + v42 . To show that it does not divide I, consider the sum of those terms in I which involve only v1 and v2 and note that this is not divisible by v12 +v22 .  Hence, I can be taken as one of the basic invariants, and j = 1 if dj = 8. 2

& & Remark. &2 C2 & is not divisible by 3. Aside from cyclic groups of prime order, these are the only known finite simple groups with this property. Now we consider the automorphisms of the twisted groups. As for the untwisted groups diagonal automorphisms and field automorphisms can be defined. Theorem 36. Let G and σ be as in this section and Gσ the subgroup of G (or Gσ ) generated by Uσ and Uσ− . Assume that σ is not the identity. Then every automorphism of Gσ is a product of an inner, a diagonal, and a field automorphism. Remark. Observe that graph automorphisms are missing. Thus the twisted groups cannot themselves be twisted, at least not in the simple way we have been considering. Sketch of proof. As in step (1) of the proof of Theorem 30, the automorphism, call it ϕ, may be normalized by an inner automorphism so that it fixes Uσ and Uσ− (in the finite case by Sylow’s theorem, in the infinite case by arguments from the theory of algebraic groups). Then it also fixes Hσ , and it permutes the Xa ’s (a simple, a ∈ Σ/R; henceforth we write Xa for Xa,σ ) and also the X−a ’s according to the same permutation, in an angle preserving manner (see step (2)) in terms of the corresponding simple system Πσ of Vσ . By checking cases one sees that the permutation is necessarily the identity: if k is finite, one need only compare the various |Xa |’s with each other, while if k is arbitrary further argument is necessary (one can, for example, check which Xa ’s are abelian and which are not, thus ruling out all possibilities except for 2 A3 , 2 E6 , and 3 D4 , and then rule out these cases (the first two together) by considering the commutator relations among the Xa ’s). As in step (4) of the proof of Theorem 30, we need only complete the proof of our theorem when Gσ is one of the groups Ga = Xa , X−a , in other words, when Gσ is one of the types A1 , 2 A2 , 2 C2 or 2 G2 (with 2 C2 (2) and 2 G2 (3) excluded, but not A1 (2), A1 (3), or 2 A2 (4)), which we henceforth assume. The case A1 having been treated in Chapter 10, we will treat only the other cases, in a sequence of steps. We write x(t, u) or x(t, u, v) for the general element of Uσ as given in Lemma 63 and d(s) for hα (s)hβ (sθ ).

44 Here,

α, β denote linear polynomials, unique up to multiplication by nonzero scalars, determining the reflecting hyperplanes corresponding to the roots α, β respectively. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(1) We have the equations d(s)x(t, u)d(t)−1 = x(s2−θ t, s1+θ u)

in 2 A2

= x(s2−2θ t, s2θ u)

in 2 C2

d(s)x(t, u, v)d(s)−1 = x(s2−3θ t, s−1+3θ u, sv)

in 2 G2 .

This follows from the definitions and Lemma 20(c). (2) Let U1 , U2 be the subgroups of Uσ obtained by setting t = 0, then also u = 0. Then Uσ ⊃ U1 ⊃ U2 = 1 is the lower central series Uσ ⊃ (Uσ , Uσ ) ⊃ (Uσ , (Uσ , Uσ )) ⊃ · · · for Uσ if the type is 2 A2 or 2 C2 , while Uσ ⊃ U1 ⊃ U2 ⊃ 1 is if the type is 2 G2 . Exercise. Prove this. (3) If the case 2 A2 (4) is excluded, then d(s)x(t, . . .)d(s)−1 = x(g(s)t, . . .), with g : k∗ → k∗ a homomorphism whose image generates k additively. Proof of (3). Consider 2 A2 . By (1) we have g(s) = s2−θ , so that g(s) = s for s ∈ kθ . Since [k : kθ ] = 2, we need only show that g takes on a value outside of kθ . Now if g doesn’t, then s2−θ = (s2−θ )θ so that s3 ∈ kθ , for all s ∈ k∗ , whence we easily conclude (the reader is asked to supply the proof) that k has at most 4  elements, a contradiction. For 2 C2 and 2 G2 the proof is similar, but easier. (4) The automorphism ϕ (of Gσ ) can be normalized by a diagonal and a field automorphism to be the identity on Uσ /U1 . Proof of (4). Since ϕ fixes Uσ , it also fixes U1 , hence acts on Uσ /U1 . Thus there is an additive isomorphism f : k → k such that ϕx(t, . . .) = x(f (t), . . .). By multiplying ϕ by a diagonal automorphism we may assume f (1) = 1. Since ϕ fixes Hσ , there is an isomorphism i : k∗ → k∗ such that ϕd(s) = d(i(s)). Combining these equations with the one in (3) we get f (g(s)t) = g(i(s))f (t) for all s ∈ k∗ , t ∈ k. Setting t = 1, we get (∗) f (g(s)) = g(i(s)), so that f (g(s)t) = f (g(s))f (t). If the case 2 A2 (4) is excluded, then f is multiplicative on k by (3), hence is an automorphism. The same conclusion, however, holds in that case also since f fixes  0 and 1 and permutes the two elements of k not in kθ . Our object now is to show that once the normalization in (4) has been attained ϕ is necessarily the identity. (5) ϕ fixes each element of U1 /U2 and U2 , and also some w ∈ Gσ which represents the nontrivial element of the Weyl group. Proof of (5). The first part easily follows from (2) and (4), then the second follows as in the proof of Theorem 33(b).  (6) If the type is 2 C2 or 2 G2 , then ϕ is the identity. Proof of (6). Consider the type 2 C2 . From the equation (∗) of (4) and the fact that f = 1, we get g(s) = g(i(s)), i.e., s2−2θ = i(s)2−2θ , and then taking the (1 + θ)-th power, s = i(s); in other words ϕ fixes every d(s). By (4) and (5), ϕx(t, u) = x(t, u + j(t)) with j an additive homomorphism. Conjugating this equation by d(s) = ϕd(s), using (1), and comparing the new equation with the old, we get j(s2−2θ t) = s2θ j(t), and on replacing s by s1+θ , j(st) = s1+2θ j(t). Choosing s = 0, 1, which is possible because 2 C2 (2) has been excluded, and replacing s by Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

11. SOME TWISTED GROUPS

113

s + 1 and by 1 and combining the three equations, we get (s + s2θ )j(t) = 0. Now s + s2θ = 0, since otherwise we would have s + s2θ = (s + s2θ )2θ , then s = s2 , contrary to the choice of s. Thus j(t) = 0. In other words ϕ fixes every element of Uσ . If the type is 2 G2 instead, the argument is similar, requiring one extra step.  Since Gσ is generated by Uσ and the element w of (5), ϕ is the identity. The preceding arugment, slightly modified, barely fails for 2 A2 , in fact fails just for the smallest case 2 A2 (4). The proof to follow, however, works in all cases. (7) If the type is 2 A2 , then ϕ is the identity. Proof of (7). Choose w as in (5) and, assuming u = 0, write wx(t, u)w−1 = xnx with x, x ∈ Uσ , n ∈ Hσ w. A simple calculation in SL3 shows that x = x(atu−1 , ·) for some a ∈ k∗ depending on w but not on t or u. (Prove this.) If now we write ϕx(t, u) = x(t, u + j(t)), apply ϕ to the above equation, and use (4) and (5), we get t¯ u−1 = t(u + j(t))−1 , so that j(t) = 0 and we may complete the proof as before.  



This completes the proof of Theorem 36.

It is also possible to determine the isomorphisms among the various Chevalley groups, both twisted and untwisted. We state the results for the finite groups, omitting the proofs. Theorem 37. (a) Among the finite simple Chevalley groups, their twisted analogues, and the alternating groups An (n ≥ 5), a complete list of isomorphisms is given as follows. (1) Those independent of k: C1 ∼ = B1 ∼ = A1 C2 ∼ = B2 D2 ∼ = A1 × A1 D3 ∼ = A3

D2 ∼ = A1 = 2 (A1 × A1 ) ∼ D3 ∼ = 2 A3 2 A1 (q 2 ) ∼ = A1 (q).

2

2

(2) Bn (q) ∼ = Cn (q) if q is even. (3) Just six other cases, of the indicated orders: A1 (4) ∼ = A1 (5) ∼ = A5 A1 (7) ∼ = A2 (2) A1 (9) ∼ = A6 A3 (2) ∼ = A8 2 A3 (4) ∼ = B2 (3)

60 168 360 20160 25920.

(b) In addition there are the following cases in which the Chevalley group just fails to be simple: The derived group of

B2 (2) ∼ = A6 G2 (2) ∼ = 2 A2 (9) 2 G2 (3) ∼ = A1 (8) 2 F4 (2).

The indices in the original group are 2, 3, 2, 2, respectively. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

360 6048 504

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Remarks. (a) The existence of the isomorphism in (1) and (2) is easy, and in (3) is proved, e.g., in Dieudonn´e [23]. There also the first case of (b), considered in the form B2 (2) ∼ S6 (symmetric group) is proved. (b) It is natural to include the simple groups An in the above comparison since they are the derived groups of the Weyl groups of type An−1 and the Weyl groups in a sense form the skeletons of the corresponding Chevalley groups. We would like to point out that the Weyl groups W (En ) are also almost simple and are related to earlier groups as follows. Proposition. We have the isomorphisms: DW (E6 ) ∼ = B2 (3) ∼ = 2 A3 (4) DW (E7 ) ∼ = C3 (2) DW (E8 )/C ∼ = D4 (2) with C the center, of order 2. Proof. The proof is similar to the proof of S6 = W (A5 ) ∼ = B2 (2) given near the beginning of Chapter 10.  Aside from the cyclic groups of prime order and the groups considered above, only 11 or 12 other finite simple groups are at present (as of May 1968) known.45 We will discuss them briefly. (a) The five Mathieu groups Mn (n = 11, 12, 22, 23, 24). These were discovered by Mathieu about a hundred years ago and put on a firm footing by Witt [65]. They arise as highly transitive permutation groups on the indicated numbers of letters. Their orders are |M11 | = 7920 = 8 · 9 · 10 · 11 |M12 | = 95040 = 8 · 9 · 10 · 11 · 12 |M22 | = 443520 = 48 · 20 · 21 · 22 |M23 | = 10200960 = 48 · 20 · 21 · 22 · 23 |M24 | = 244823040 = 48 · 20 · 21 · 22 · 23 · 24 (b) The first Janko group J1 discovered by Janko [37] about five years ago. It is a subgroup of G2 (11) and can be represented as a doubly transitive permutation group on 266 letters. Its order is |J1 | = 175560 = 11(11 + 1)(113 − 1) = 19 · 20 · 21 · 22 = 55 · 56 · 57 The remaining groups were all discovered last fall, more or less. (c) The groups J2 and J2 1/2 of Janko. The existence of J2 was put on a firm basis first by Hall and Wales using a machine, and then by Tits in terms of a “geometry.” It has a subgroup of index 100 isomorphic to DG2 (2) ∼ = 2 A2 (9), and is itself of index 416 in G2 (4). The group J2 1/2 has not yet been put on a firm basis, and it appears that it will take a great deal of work to do so (because it 45 Aside

from the five Mathieu groups discovered in the 1860s, there are 21 more so called sporadic simple groups found between 1965 and 1975. The Classification of Simple Groups theorem claims that cyclic groups, alternating groups, groups of Lie type, and 26 sporadic groups exhaust the list of all finite simple groups. The Wikipedia article List of finite simple groups may serve as good entry point; it also contains useful references. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

11. SOME TWISTED GROUPS

115

does not seem to have any “large” subgroups), but the evidence for its existence is overwhelming. The orders are: |J2 | = 604800 & & &J2 1/2 & = 50232960 (d) The group H of D. Higman and Sims, and the group H  of G. Higman. The first group contains M22 as a subgroup of index 100 and was constructed in terms of the automorphism group of a graph with 100 vertices whose existence depends on the properties of Steiner systems. Inspired by this construction, G. Higman then constructed his own group in terms of a very special geometry invented for the occasion. The two groups have the same order, and everyone seems to feel that they are isomorphic, but no one has yet proved this. The order is: |H| = |H  | = 44352000 (e) The (latest) group S of Suzuki. This contains G2 (4) as a subgroup of index 1782, and is constructed in terms of a graph whose existence depends on the imbedding J2 ⊂ G2 (4). It possesses an involutory automorphism whose set of fixed points is exactly J2 . Its order is: |S| = 448345497600 (f) The group M of McLaughlin. This group is constructed in terms of a graph and contains 2 A3 (9) as a subgroup of index 275. Its order is |M | = 898128000 Theorem 38. Among all the finite simple groups (i.e., all that are currently known), the only coincidences in the orders which do not come from isomorphisms are: (a) Bn (q) and Cn (q) for n ≥ 3 and q odd. (b) A2 (4) and A3 (2) ∼ = A8 . (c) H and H  if they aren’t isomorphic. That the groups in (a) have the same order and are not isomorphic has been proved earlier. The orders in (b) are both equal to 20160 by Theorem 25, and the groups are not isomorphic since relative to the normalizer B of a 2-Sylow subgroup the first group has six double cosets and the second has 24. The proof that (a), (b) and (c) represent the only possibilities depends on an exhaustive analysis of the group orders which can not be undertaken here.

Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

https://doi.org/10.1090//ulect/066/13

CHAPTER 12

Representations In this chapter we consider the irreducible representations of the infinite Chevalley groups. As we shall see, here the theory is quite complete. All representations are assumed to be finite-dimensional and the standard terminology is used. In particular 1 must act as the identity, and the trivial 0-dimensional (but not the trivial 1-dimensional) representation is excluded from the list of irreducible representations. We start with a general lemma. Lemma 68. Let K be an algebraically closed field, B and C associative algebras with 1 over K, and A = B ⊗ C. (a) If (β, V ) and (γ, W ) are (finite-dimensional) irreducible modules for B and C, then (α, U ) = (β ⊗ γ, V ⊗ W ) is one for A. (b) Conversely, every irreducible A-module (α, U ) is realizable, uniquely, as a tensor product as in (a). Proof. (a) By Burnside’s Theorem (see, e.g., Jacobson [36]), βB = End V and γC = End W , whence αA = End U and (α, U ) is irreducible. (b) Let V be an irreducible B-submodule of U . Such exist since U is finitedimensional. Let L be the space of B-homomorphisms of V into U . This is nonzero and is a C-module under the rule cl = α(c) ◦ l. (Check this.) Let (γ, W ) be an irreducible submodule. The map ϕ : V ⊗ W → U defined by v ⊗ f → f (v) is easily checked to be an A-homomorphism. V ⊗ W is irreducible by (a), and U is by assumption. Hence by Schur’s Lemma (see loc. cit.) ϕ is an isomorphism. If α = β  ⊗ γ  is a second decomposition of the required form, then restriction to B yields β ⊗ 1 ∼ = β  ⊗ 1, i.e., multiples of β and β  are isomorphic, so that by the Jordan-H¨ older or Krull-Schmidt theorems β and β  are also. Similarly γ and γ  are isomorphic, which proves the uniqueness in (b).  Corollary.

(a) If K is an algebraically closed field and G = Gi is a direct product of a finite number of groups, then the tensor product V of irreducible KGi modules Vi is an irreducible KG-module, and every irreducible KG-module is uniquely realizable in this way. + (b) Similarly for a direct sum L = Li of Lie algebras over K.

Proof. We apply Lemma 68, extended to several factors, in (a) to group algebras, in (b) to enveloping algebras.  Exercise. If the direct product above is one of algebraic groups over K (of topological groups, of Lie groups, . . .), then V is rational (continuous, analytic, . . .) if and only if each Vi is. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 117

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Remark. If we are interested in the irreducible representations of a Chevalley group G, we may as well assume it is universal. The corollary then implies that we may also assume that G (i.e., that Σ) is indecomposable. This we will do whenever it is convenient. Now we take up the study of rational representations for Chevalley groups over algebraically closed fields viewed as algebraic groups. In such representations the coordinates of the representative matrix are required to be rational functions of the original coordinates. Whether this requirement is to be taken locally (e.g., as in the poof of Theorem 7, Corollary 1) or globally is immaterial, in view of the following result. Lemma 69. Let G be a Chevalley group viewed as an algebraic group as above, and f : G → k a function. Then the following conditions are equivalent. (a) f is expressible as a rational function locally. (b) f is expressible as a rational function globally. (c) f is expressible as a polynomial. Proof. It will be enough to show that (a) implies (c). Let A be the algebra of polynomial functions on G. By assumption there exists an open covering {Ui } of G, which may be taken finite by the maximal condition on the open subsets of G (which holds by Hilbert’s basis theorem in A), and elements gi , hi in A such that together, by f = gi /hi and hi = 0 on Ui for all i. Since the hi don’t all vanish Hilbert’s Nullstellensatz there exist elements a in A such that 1 = ai hi on G. i # = U ; it is nonempty, in fact dense, since G is irreducible. On U Let U 0 i  0 we have ai gi , a polynomial, hence by density also on each Ui and on G, f = ai f hi = as required.  Presently we will need the following result. Lemma 70. The algebra A of polynomial functions on G is integrally closed (in its quotient field). Proof. We observe first that A is an integral domain (since G is irreducible as an algebraic set, the polynomial ideal defining it is prime), so that it really has a quotient field. Assume f = p1 /p2 (pi ∈ A) is integral over A: f n + a1 f n−1 + · · · + an = 0 for some ai ∈ A. On restriction to the open set U − HU  of G the  pi and . Since ai become, by Theorem 7(b), polynomials in the coordinates tα , ti , t−1 i such polynomials form a unique factorization domain, we see by the above equation that f itself is such a polynomial, on U − HU . The same being true on each of the translations of U − HU by elements of G, we conclude that f is a polynomial on G, i.e., f ∈ A, by Lemma 69.  Two more lemmas and then the main theorem. Lemma 71. The rational characters of H (homomorphisms into k∗ ) are just the elements of the lattice L generated by the global weights of the representation defining G. Proof. Let λ be a character. Then it is a polynomial in the diagonal elements of H (written as a group of diagonal matrices), i.e., a linear combination of elements of L. Being multiplicative, it equals some element of L. (Prove this.) Conversely, if λ ∈ L, then λ is a power product of weights in the representation defining G, Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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and all exponents may be taken positive since the product of the latter weights (as functions) is 1, so that λ is a polynomial on H.  Now for any rational G-module V we may define the weights λ and the corresponding weight spaces Vλ , relative to H, the obvious way. Lemma 72. Let V be a rational G-module, λ a weight, v an element of Vλ , and α a root. Then there exist vectors vi ∈ Vλ+iα (i = 1, 2, . . .) so that xα (t)v = v + ti vi for all t ∈ k. Proof. Since V is  rational and t → xα (t) is an isomorphism, xα (t) is a polyti vi . If we apply h to this equation and compare the nomial in t: xα (t)v = result with the equation got by replacing xα (t) by hxα (t)h−1 = xα (α(h)t), we get  vi ∈ Vλ+iα . Setting t = 0, we get v = v0 , whence the lemma. Theorem 39 (Compare with Theorem 3). Let G be a Chevalley group over an algebraically closed field k (i.e., a semisimple algebraic group over k), and assume the notations as above. (a) Every nonzero rational G-module V contains a nonzero element v + which belongs to some weight λ ∈ L and is fixed by all x ∈ U . (b) Assume V = kGv + with v + as in (a). Then V = kU − v + . Further 1, every weight μ on V has the form λ − α (α positive root), dim Vλ =  and V = Vμ . (c) In (a) λ, α ∈ Z+ for every positive root α. (d) If V is irreducible, then the weight λ (the “highest weight”) and the line kv + of (a) are uniquely determined. (e) Given any character λ on H satisfying (c), there exists a unique irreducible rational G-module V in which λ is realized as in (a). Proof. (a) The proof is the same as that of Theorem 3(a) with Lemma 72 in place of Lemma 11. (b) Since U − B is dense in G (Theorem 7) and V is rational, any linear function on V which vanishes on U − Bv + also vanishes on Gv + . Thus V = kU − Bv + = kU − v + . The other assertions of (b) follow from this equation and Lemma 72. (c) wα λ is a weight on V (with wα v + a corresponding weight vector). Since wα λ = λ − λ, α α, it follows from (b) that λ, α ∈ Z+ . (d) This follows from the second and third parts of (b). (e) We will use the correspondence between local weights (on L) and global weights (on G) (see page 39). Let λ be as in (e). By Lemma 71, λ ∈ L. Let λ also denote the corresponding weight on L, so that λ(Hα ) = λ, α ∈ Z+ for all α > 0. Let (ρ, V  ) be an irreducible L-module with λ as its highest weight, v + a corresponding weight vector, and G a corresponding Chevalley group over k (constructed from ρL and some choice of lattice M in V  ). Since λ is in the lattice generated by the weights of the representation of L used to construct G, it follows (Theorem 7, Corollary 1) that there exists a rational homomorphism ϕ : G → G such that xα (t) → xα (t) or 1 for all α and t, in the usual notation. The resulting representation of G on V  need not be irreducible (and its representation class may vary with the choice of M ), but at least it contains the vector v + which is of weight λ and is fixed by every x ∈ U . Let V  be the submodule of V  generated by v + and Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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V  a maximal submodule of V  . (V  is in fact unique as follows from the equation V  = kU − v + of (b). Check this.) The G-module V = V  /V  meets the existence requirements of (e). For the uniqueness, let Vi , vi+ (i = 1, 2) satisfy the conditions on V, v + in (e). Let v + = (v1+ , v2+ ) ∈ V1 ⊕ V2 , and V = kGv + . Then Vλ = kv + by / V . Consider the G-homomorphism p1 : V → V1 , projection on (b), so that v2+ ∈ the first factor. Since v1+ generates V1 , p1 is onto. Since also ker p1 ⊆ V2 ∩ V , which / V , it follows that p1 is an isomorphism. is zero because V2 is irreducible and v2+ ∈ Thus V is isomorphic to V1 , and similarly to V2 , so that V1 and V2 are isomorphic, as required.  A complement. If char k = 0, then in the existence proof above V  is itself irreducible, i.e., V  = V  and V  = 0. In other words, if the Chevalley group G is constructed from an irreducible L-module V and a field k of characteristic 0, then as a linear group it is irreducible. Proof. Recall that V was originally an LC -module (irreducible by assumption), and that a lattice M as in Theorem 2, Corollary 1 was then used to shift the coefficients to k. Clearly VQ is irreducible relative to LQ . It follows that Vk is irreducible relative to Lk : otherwise there would be a proper invariant subspace V1 , then some nonzero v ∈ V1 such that Xα v = 0 excluding kv + since kv + ∩ VQ = 0,  for all α > 0, so that writing v = ti mi (mi ∈ M, ti ∈ k and linearly independent over Q) and  choosing α so that Xα mi = 0 for some i, we would arrive at the contradiction ti Xα mi = 0. Since we can recover each Xα from G by using xα (t) = 1 + tXα + · · · for several values of t and the Xα ’s generate L, we conclude  that Vk is irreducible for G. In contrast to the case just considered, if char k = 0, then V  = V  and V  = 0 in general and the exact situation is not all understood, except in a few scattered cases (types A1 , A2 , B2 or when char k is large “compared” to λ). However, the following is true. Exercise. (a) For the lattice M of Theorem 2, Corollary 1 (with V there assumed to be irreducible) assume that Cv + ∩ M is prescribed. Prove that there is a unique minimal choice for M (contained in all others) and a unique maximal choice. Assume now as in the complement, except char k = 0. (b) If M is maximal, then V  = 0, i.e., V  is irreducible. (c) If M is minimal, then V  = V  . Example. If L is of type A1 , char k = 2, and the adjoint representation is used, then (b) holds for Mmax = X, H/2, Y  and (c) holds for Mmin = X, H, Y , but not vice versa. The proof given above for the existence in Theorem 39(e) brings out the connection between the representations of G and those of L and shows that every irreducible rationals representation of a Chevalley group in characteristic p = 0 can be constructed by the reduction mod p of a corresponding representation of a group in characteristic 0. It depends, however, on the existence of representations of L, which we have not proved here, thus in its entirety is very long. We shall now develop an alternate, more intrinsic proof. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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We start with the connection between a G-module V and its dual V ∗ , on which G acts by the rule (xf )(v) = f (x−1 v) for all x ∈ G, f ∈ V ∗ , and v ∈ V . We recall that w0 is the element of the Weyl group which makes all positive roots negative. Lemma 73. Let V be an irreducible rational G-module, λ its highest weight, v + a corresponding weight vector, λ∗ = −w0 λ, and f + the element of V ∗ defined thus: if we write v − = w0 v + and v ∈ V as cv − + terms of other (hence higher) weights, then f + (v) = c. Then λ∗ and f + are highest weight and highest weight vector for V ∗. Proof. In the definiton of f + we have used the fact that dim Vw0 λ = dim Vλ = 1. Here, and also in similar situations later, we extend λ to B by the rule λ(b) = λ(h) if b = uh (u ∈ U, h ∈ H), and similarly for λ∗ . If we write v ∈ V as in the lemma and use Lemma 72, we see that bv = c(w0 λ)(h)v − + higher terms. Since c = f + (v) and (w0 λ)(h) = λ∗ (b−1 ), we have f + (bv) = λ∗ (b−1 )f + (v). On replacing b by b−1 we get bf + = λ∗ (b)f + , as required.  Theorem 40. For λ ∈ L let Aλ be the space of polynomial functions a on G such that a(yb) = a(y)λ(b) for all y ∈ G, b ∈ B, made into a G-module in the obvious way. (a) If V, λ, v + are as in Lemma 73, then the map ϕ : V ∗ → Aλ defined by (ϕf )(x) = f (xv + ) for f ∈ V ∗ and x ∈ G is a G-isomorphism into. (b) Conversely, if λ is such that Aλ = 0, then Aλ contains a unique irreducible G-submodule. The latter is finite-dimensional and rational and its highest weight is λ∗ . Proof. (a) The points to be checked here will be left as an exercise. (b) We observe first that as a G-module Aλ is locally finite-dimensional (in fact, it is finite-dimensional, but we shall not prove this), since the set of polynomials of a given degree is. Thus there exist irreducible submodules and all of them are finitedimensional and rational. Let μ be the highest weight of any one of them and a+ a corresponding nonzero weight vector. We have (∗) a+ (bxb ) = μ(b−1 )a+ (x)λ(b ) for all x ∈ G, b, b ∈ B. Since Bw0 B is dense in G, a+ (w0 ) = 0. Since also a+ (bw0 ) = a+ (w0 · w0−1 bw0 ), we get from the above equation that μ(b−1 ) = λ(w0−1 bw0 ), so that μ = λ∗ . Since μ is uniquely determined by λ, the function a+ is determined by its value at w0 by (∗) with x = w0 and the density of Bw0 B in G, proving the uniqueness in (b).  Remarks. (a) In characteristic 0 it easily follows from the theorem of complete reducibility that Aλ itself is irreducible. (b) The representation of G on Aλ is, in the context of polynomial representations, the one induced by the character λ on B. The fact that it contains a representation of highest weight λ∗ is, in view of Theorem 39(a), a form of Frobenius reciprocity. Lemma 74. Let f + be as in Lemma 73 and a+ = ϕf + with ϕ as in Theorem 40(a) so that xv + = a+ (x)w0 v + + higher terms. Let Wλ be the stabilizer of λ in W , and for w ∈ Wλ assume that the corresponding representative w ∈ G has been chosen so that wv + = v + . Then if x ∈ G is written uhw0 wu1 (see Theorem 4 ) we Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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have

® +

a (x) =

λ∗ (h−1 ), 0,

if w ∈ Wλ , otherwise

Proof. A choice for w ∈ G as above is always possible: If λ = 0, then V is trivial since G = DG, while if λ = 0, when wv + has weight wλ = λ, hence is a multiple of v + , so that by modifying it by a suitable element of H we can achieve wv + = v + . From the definitions and Lemma 72 we have a+ (x)w0 v + = hw0 wv + . If w ∈ Wλ , then a+ (x) = λ∗ (h−1 ) by the choice of w, while if a+ (x) = 0, then w  fixes kv + by the equation so that w ∈ Wλ . This brings us to the Second proof of the existential part of Theorem 39(e). Proof. Let λ be as in Theorem 39(c). It will be enough to prove that the function defined by the last equation of Lemma 74 is rational on G. The existence will then follow from Theorem 40(b) with λ∗ in place of λ. By Lemma 70 any power of this function will do, so that by Lemma 74 it will be enough to construct an irreducible representation whose highest weight is some positive power (positive multiple if we write characters on H additively) of λ. This we will do, using the following interesting result. Lemma 75 (Chevalley). Let G be a linear algebraic group and P a closed subgroup. Then there exists a rational G-module V and a line L in V whose stabilizer in G is P . Proof. Let A be the algebra of polynomials in the matrix entries and I the ideal defining P . By Hilbert’s basis theorem I is generated by a finite number of its elements, so that there exists a finite-dimensional G-invariant subspace B of A such that B ∩ I, say C, generates I. For x ∈ G we have the following equivalent conditions: x ∈ " P ; f (x−1 y) = 0 for all f ∈ I, y ∈ P ; xI ⊆ I; xC ⊆ C. If now c = dim C, V = c B, v is the product in V of a basis for C, and L = kv, it follows that the stabilizer of L is exactly P .  We resume the proof of existence. Let Π = {α1 , α2 , . . . , αl } be the set of simple roots. For i = 1, 2, . . . , l let Pi be the parabolic subgroup of G corresponding to Π − {αi } (see Lemma 30), Li = kvi the corresponding line of Lemma 75, μi the corresponding rational character on Pi , hence also on B and H, and Vi = kGvi . If j = i, then wj is represented in Pi , so that wj μi = μi and μi , αj  = 0. Since wi does not fix Li , by choice, it follows from parts (b) and (c) of Theorem 39 say di . If now applied to Vi that μi , αi  is a positive integer,  λ is as before so that that dλ =

ei μi with d = di and ei = ci d/di . If λ, αi  = ci ∈ Z+ , it follows

viei is a vector of weight dλ for B, so we form the tensor product Viei , then that we may extract an irreducible component whose highest weight is dλ, and thus complete our second existence proof.  Remark. We are indebted to G. D. Mostow for the proof just given. The extra problems that arise when char k = 0 are compensated for by the fact that only a finite number of representations has to be considered in this case, as we shall now see. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Lemma 76. Assume char k = p = 0. Let Fr (for Frobenius) denote the operation of replacing the matrix entries of the elements of G by their p-th powers. If ρ is an irreducible rational representation of G, then so is ρ ◦ Fr. If the highest weight of ρ is λ, that of ρ ◦ Fr is pλ. Proof. Exercise.



Theorem 41. Assume that G above is universal (i.e., G is a simply connected algebraic group), and that char k = p = 0. Let R be the set of pl irreducible representations of G for which the highest weight λ satisfies 0 ≤ λ, αi  ≤ p − 1 (αi simple). Then every irreducible representation of G can be written uniquely

m j j=0 ρj ◦ Fr (ρj ∈ R). Sketch of proof. We observe first that since G is universal L = L1 , so that all λ’s with λ, αi  ∈ Z+ occur as highest weights, in particular those used to define

R. Consider ρ = ρj ◦ Frj . Let λj be the highest weight of ρj . The product of thecorresponding weight vectors yields for ρ a highest weight vector of weight λ = pj λj , by Lemma 76. If we vary the ρj ’s in R, we obtain, in view of the uniqueness of the expansion of a number in the scale of p, each possible highest weight λ exactly once. Thus to prove the theorem we need only show that each ρ above is irreducible. The proof of this fact depends eventually on the linear independence of the distinct automorphisms Frj (j = 0, 1, . . .) of k. We omit the details, referring the reader to Steinberg [60] or to Cartier [13]).  Corollary. Assume that one of the special situations of Theorem 28 holds. Let Rl (resp. Rs ) be the subsets of R defined by λ, αi  = 0 for all i such that αi is long (resp. short). Then every element of R can be written uniquely ρl ⊗ ρs with ρl ∈ Rl and ρs ∈ Rs . Proof. Given ρ ∈ R, write the corresponding highest weight λ as λl + λs so that the corresponding irreducible representations ρl and ρs are in Rl and Rs . We have to show that ρl ⊗ ρs is irreducible. If we define ϕ as in Theorem 28 but with G and G∗ interchanged, and set ρ∗l = ϕ ◦ ρl , ρ∗s = ϕ ◦ ρs , we have to show that the representation ρ∗l ⊗ ρ∗s of G∗ is irreducible. Since the corresponding highest weights satisfy ® λl , α , if α is short ∗ ∗ λl , α  = 0, if not, ® p λs , α , if α is long, λ∗s , α∗  = 0, if not, we see that λ∗l and λ∗s /p correspond to elements of R∗ , so that the corollary follows  from Theorem 41 applied to G∗ . Examples. (a) SL2 . Here there are p representations ρi (i = 0, 1, . . . , p − 1) in R, the i-th being realized on the space of polynomials homogeneous of degree i over k2 . (b) Sp4 , p = 2. Here there are 4 representations in R. If ϕ is the graph automorphism of Theorem 28 then Rl ◦ ϕ = Rs so that by the above corollary, these 4 are, in terms of the defining representation ρ, just 1 (trivial), ρ, ρ ◦ ϕ, and ρ ⊗ (ρ ◦ ϕ). Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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The results we have obtained can easily be extended to the case that k is infinite (but perhaps not algebraically closed). We consider representations on vector spaces over K, some algebraically closed field containing k, and call them rational if the coordinates of the image are polynomial functions over K in the coordinates of the source. The preceding theory is then applicable almost word for word because of the following two facts both coming from the denseness of G in GK (this is G with k extended to K). (a) Every irreducible polynomial representation of G extends uniquely to one of GK . (b) On restriction to G every irreducible rational representation of GK remains irreducible. Exercise. Prove (a) and (b). The structure of arbitrary irreducible representations is given in terms of the polynomial ones by the following general theorem. Given an isomorphism ϕ of k into K, we shall also write ϕ for the natural isomorphism of G onto the group ϕG obtained from G by replacing k by ϕk. Theorem 42 (Borel, Tits). Let G be an indecomposable universal Chevalley group over an infinite field k, and let σ be an arbitrary (not necessarily rational) irreducible representation of G on a finite-dimensional vector space V over an algebraically closed field K. Assume that σ is nontrivial. Then there exist finitely many irreducible rational representations isomorphisms ϕi of k into K and corresponding

of ρi of ϕi G over K such that σ = i ρi ◦ ϕi . Remarks. (a) As a corollary we see that k is necessarily imbeddable as a subfield of K. In other words, if k and K are such that no such imbedding exists, e.g., if char k = char K, then every irreducible representation of G on a finitedimensional vector space over K is necessarily trivial. (Deduce that the same is true even if the representation is not irreducible.) If k is finite, these statements are, of course, false. (b) The theorem can be completed by statements concerning the uniqueness of the decomposition and the condition for irreducibility if the factors are prescribed. Since these statements are a bit complicated we shall omit them. (c) The theorem was conjectured by us in [60] The proof to follow is based on an as yet unpublished paper by A. Borel and J. Tits46 which results of a more general character are considered. Lemma 77. Let G, G be indecomposable Chevalley groups over fields k, k with k infinite and k algebraically closed, and σ : G → G a homomorphism such that σG is dense in G . (a) There exists an isomorphism ϕ of k into k and a rational homomorphism ρ of ϕG into G such that σ = ρ ◦ ϕ. (b) If G is universal, then ρ can be lifted, uniquely, to the universal covering group of G . Proof. (a) If the reader will examine the proof of Theorem 31 he will observe that what is shown there is that σ can be normalized so that σxα (t) = 46 See

Borel and Tits [7]. ()

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xα (α ϕ(t)q(α) ) with α → α an angle-preserving map of Σ on Σ , α = ±1, ϕ an isomorphism of k onto k , and q(α) = 1 on p. Since we are assuming only that σG is dense in G , not that σG = G , the proof of the corresponding result in the present case is somewhat harder. However, the main ideas are quite similar. We omit the proof. From the above equation and the corresponding ones on H, it follows from Theorem 7 that σ has the form of (a). (b) From these equations we see also, e.g., by considering the relations (A), (B), (C) of Theorem 8, that ρ can be lifted to any covering of G , uniquely since G = DG.  Proof of Theorem 42. Let A = σG, the smallest algebraic subgroup of GL(V ) containing σG. We claim A is a connected semisimple group, hence a Chevalley group. As in the proof of Theorem 30, step (12), σU is connected, and similarly for σU − , so that A, being generated by these groups, is also. Let R be a connected solvable normal subgroup of A. By the Lie-Kolchin Theorem R has weights on V , finite in number. A permutes the corresponding weight spaces, and, being connected, fixes them all. Since V is irreducible, there is only one such space and it is all of V , so that R consists of scalars, of determinant 1 since A = DA, so that R is finite. Since R is connected, R = 1, so that A is semisimple, as claimed. group of A written as a product of its Let A1 = Ai1 be the universal covering Ai0 the corresponding factorization of the indecomposable components, A0 = adjoint group, and α, β, γ = γi the corresponding natural maps as shown: A1 =



α

δ σ

G

Ai1

γ=

A

βσ



γi

β A0 =



Ai0

By Lemma 77 we can lift βσ componentwise to get a homomorphism δ : G → A1 of the form δ(x) = i ϕi (x) with each ϕi an isomorphism of k into K and i a rational homomorphism of ϕi G into Ai1 . We have αδ = σ since otherwise we would have a homomorphism of G into the center of A. By Lemma 68, Corollary (a), α

α with αi an irreducible rational can be factored i i

representation of Ai1 . On αi i ϕi = i ρi ϕi , as required.  setting ρi = αi i , we see that σ = αδ = Corollary. (a) Every absolutely irreducible real representation of a real Chevalley group G is rational. (b) Every holomorphic irreducible representation of a semisimple complex Lie group is rational. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(c) Every continuous irreducible representation of a simply connected semisimple complex Lie group is the tensor product of a holomorphic one and an antiholomorphic one. Proof. (a) If G is universal, this follows from the theorem and the fact that the only isomorphism of R into R is the identity. The transition to the nonuniversal case is an easy exercise. (b) The proof is similar to that of (a). (c) The only continuous isomorphisms of C into C are the identity and complex conjugation.  Exercise. Prove that the word “absolutely” in (a) and the words “simply connected” in (c) may not be removed. Now we shall touch briefly on some additional results. Characters. As is customary in representation theory, the characters (i.e., the traces of the representative matrices) play a vital role. We state the principal results in the form of an exercise. Exercise. (a) Prove that two irreducible rational G-modules are isomorphic if and only if their characters are equal. (Consider the characters on H.) (b) Assume that char k = 0 and that the theorem of complete reducibility has been proved in this case. Prove (a) for representations which need not be irreducible. (c) Assume char k = 0. Prove Weyl’s formulas: Let V, λ be as in Theorem 39(e), χ the corresponding character, and δ one-half the sum of the positive roots, a  character on H. Set Sλ = w∈W (det w) · w(λ + δ), a sum of functions on H. Then (1) χ(h) = Sλ (h)/S0 (h) at all h ∈ H where S0 (h) = 0. (2) dim V = α>0 λ + δ, α / δ, α. (Hint: Use the corresponding formulas for Lie algebras (see, e.g., Jacobson, Lie Algebras [35]) and the complement to Theorem 39.) Remark. The formula (1) determines χ uniquely since it turns out that the elements of G which are conjugate to those elements of H for which S0 = 0 form a dense open set in G. The unitarian trick. The basic results about the irreducible complex representations of a compact semisimple Lie group K, i.e., a maximal compact subgroup of a complex Chevalley group G as in Chapter 8, can be deduced from those of G because of the following important fact: (∗) K is Zariski-dense in G. Because of Lemmas 43(b) and 45 (K is generated by the group ϕ2 SU2 ) this comes down to the fact that SU2 is Zariski-dense in SL2 (C), whose proof is an easy exercise. By (∗) the rational irreducible representations of G remain distinct and irreducible on restriction to K. That a complete set of continuous representations of K is so obtained then follows from the fact that the corresponding characters form a complete set of continuous class functions on K. The proof of this uses the formula for Haar measure on K and the orthogonality and completeness properties of complex exponentials, and yields as a by-product Weyl’s character formula itself. This is how Weyl proved his formula in [64] and is still the best way. The theorem of Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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complete reducibility can be proved as follows. Given any rational representation space V for G and an invariant subspace V  , we can, by averaging over K, relative to Haar measure, any projection of V onto V  and taking the kernel of the result, get a complementary subspace invariant under K, thus also invariant under G because of (∗). It is then not difficult to replace the complex field by any field of characteristic 0. Invariant bilinear form. G denotes an indecomposable infinite Chevalley group, V an irreducible rational G-module, and λ its highest weight. Lemma 78. The following conditions are equivalent. (a) There exists on V a (nonzero) invariant bilinear form. (b) V and its dual V ∗ are isomorphic. (c) −w0 λ = λ. Proof. Exercise (see Lemma 73).



Exercise. Prove that −w0 is the identity for all simple types except An (n ≥ 2), D2n+1 , E6 and for these types it comes from an involutory automorphism of the Dynkin diagram. (Hint: for all of the unlisted cases except for D2n the Dynkin diagram has no symmetry. Exercise. If there exists an invariant bilinear form on V , then it is unique up to multiplication by a scalar and is either symmetric or skew-symmetric. (Hint: use Schur’s Lemma.) Lemma 79. Let h = hα (−1), the product over the positive roots. (a) h is in the center of G and h2 = 1. (b) If V possesses an invariant bilinear form then it is symmetric if λ(h) = 1, skew-symmetric if λ(h) = −1. Proof. (a) Since hα (−1) = h−α (−1) (check this), h is fixed by all elements of W . This implies that h is in the center, as easily follows from Theorem 4 . Since hα (−1)2 = hα (1) = 1, we have h2 = 1. (b) We have an isomorphism ϕ : V → V ∗ , v + → f + with v + and f + as in Lemma 73; the corresponding bilinear form on V is given by (v, v  ) = (ϕv)(v  ). It follows that (xv + , yv + ) = f + (x−1 yv + ) for all x, y ∈ G. Thus (v + , w0 v + ) = f + (w0 v + ) = 0 by the definition of f + , and (w0 v + , v + ) = f (w0−1 v + ). If w0 = wα wβ wγ · · · is a minimal product of simple reflections in W , then for definiteness we pick w0 = wα (1)wβ (1) · · · in G, so that w0−1 = · · · wγ (−1)wβ (−1)wα (−1). We have wα (−1) = wα (1)hα (1), and similarly for β, γ, . . .. Substituting into the expression for w0−1 and bringing all the h’s to the right, by repeated conjugation by w’s, we get to the right h by Appendix II(25) and to the left · · · wγ (1)wβ (1)wα (1) which is just w0 by a lemma to be proved in the last section. Thus w0−1 = w0 h, and (w0 v + , v + ) becomes λ(h)f + (w0 v + ) = λ(h)(v + , w0 v + ), as required.  Observation. h as in Lemma 79 is 1 in each of the following cases, since the center is of odd order. (a) G is adjoint. (b) char k = 2. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(c) G is of type A2n , E6 , E8 , F4 , G2 . Exercise. In the remaining cases find h, as a product simple roots.



hα (−1)nα over the

Exercise. For every type of G (An , Bn , . . .) find all the λ for which Vλ has a bilinear invariant, and then all the λ for which the invariant is symmetric, and all the λ for which it is skew. Example. SL2 . For every V there is an invariant bilinear form. Assume char k = 0, so that for each i = 1, 2, 3, . . . there is exactly one V of dimension i, viz., the space of polynomials homogeneous of degree i−1. Then the invariant form is symmetric if i is odd, skew-symmetric if i is even. Invariant Hermitian forms. Assume now that G is complex, σ is the automorphism of Theorem 16, K = Gσ is the corresponding maximal compact subgroup, V and v + are as before, and f : G → C is defined by xv + = f (x)v + + terms of other weights. (a) Prove that f (σx−1 ) = f (x). (First prove it on U − HU , then use the density of U − HU in G.) (b) Prove that there exists a unique form (·, ·) from V × V to C which is linear in the second position, conjugate linear in the first, and satisfies (xv + , yv + ) = f (σx−1 · y), and that this form is Hermitian. (c) Prove that (·, ·) is positive definite and invariant under K. Dimensions. Assume now that G is a Chevalley group over an infinite field k, that V and λ are as before, and that UZ is the universal algebra of Theorem 2, written in the form UZ − UZ ◦ UZ+ of page 15. (a) Prove that there exists an antiautomorphism σ of UZ such that σXα = X−α and σHα = Hα for all α. (b) Define a bilinear form (u, u ) from UZ to UZ◦ thus: write σu.u in the above form and then set every Xα = 0. Prove that this form is symmetric. (c) Now define a bilinear form from UZ to Z thus: (·, ·)λ = λ◦(·, ·) (interpreting λ as a linear form on H such that λ(Hα ) ∈ Z+ for all α > 0). Assuming now that this form is reduced modulo the characteristic of k, prove that its rank is just the dimension of V .

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https://doi.org/10.1090//ulect/066/14

CHAPTER 13

Representations Continued In this chapter the irreducible representations of characteristic p (the characteristic of the base field k) of the finite Chevalley groups and their twisted analogues will be considered. The main result is as follows. Theorem 43. Let G be a finite universal Chevalley group or one of its twisted analogues considered as in Chapter 11 as the set of fixed points of an automorphism of the form xα (t) → xρα (±tq(α) ). Then the α simple q(α) irreducible polynomial representations of the including algebraic group (got by extending the base field k to its algebraic closure) for which the highest weights λ satisfy 47 0 ≤ λ, α ≤ q(α) − 1 for all simple α remain irreducible and distinct on restriction to G and form a complete set. By Theorem 41 we also have a tensor product theorem with the product ∞ 0 n−1 suitably truncated, for example, to 0 if G is a Chevalley group over a field of pn elements. Exercise. Deduce from Theorem 43 the nature of the truncations for the various twisted groups. Instead of proving the above results (see Steinberg [60]), which would take too long, we shall give an a priori development, similar to the one of the last section. We start with a group of twisted rank 1 (i.e., of type A1 , 2 A2 , 2 C2 or 2 G2 , the degenerate cases A1 (2), A1 (3), . . . not being excluded). The subscript σ on Gσ , Wσ , . . . will henceforth be omitted. We write X, Y, w, K for Xa (a > 0), X−a , the nontrivial element of the Weyl group realized in G, and an algebraically closed field of characteristic p. We observe that U = X and U − = Y in the present case. In addition, X will denote the sum of the elements of X in KG. Definition. An element v in a KG-module V is said to be a highest weight vector if it is nonzero and satisfies (a) xv = v for all x ∈ U . (b) hv = λ(h)v for all h ∈ H and some character λ on H. (c) Xwv = μv for some μ ∈ K. The couple (λ, μ) is called the corresponding weight. Remark. The refinement (c) of the usual definition is due to C. W. Curtis, [21, 22]. That such a refinement is needed is already seen in the simplest case G = SL2 (p). Here there are p representations realized on the spaces of homogeneous polynomials of degrees i = 0, 1, . . . , p − 1, with the highest weight in the usual sense being iλ1 . Since the group H is cyclic of order p − 1, the weights 0 and (p − 1)λ1 47 For

a non-twisted Chevalley group we have q(α) = q = |k| for all roots α. ()

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are identical on H, hence do not distinguish the corresponding representations from each other. Theorem 44. Let G (of rank 1) and the notations be as above. (a) Every nonzero KG-module V contains a highest weight vector v. For such a v we have KGv = KY v = KU − v. (b) If V is irreducible, then it determines Kv as the unique line of V fixed by U , hence it also determines the corresponding highest weight. (c) Two irreducible KG-modules are isomorphic if and only if their highest weights are equal. The proof depends on the following two lemmas. Lemma 80. Let Y and K be as above, or, more generally, let Y be any finite p-group and K any field of characteristic p. (a) Every nonzero KY -module V contains nonzero vectors invariant under Y. (b) Every irreducible KY -module is trivial.   (c) rad KY = { c(y)y : c(y) = 0} is the unique maximal (one-sided or two-sided) ideal of KY . It is nilpotent. (d) KY is the unique minimal ideal of KY . Proof. (a) By induction on |Y |. Assume |Y | > 1. Since Y is a p-group, it has a normal subgroup Y1 of index p, and the subspace V1 of invariants of Y1 on V is nonzero by the inductive assumption. Choose y ∈ Y to generate Y /Y1 , and v ∈ V1 nonzero. Then (1 − y)p v = 0. Now choose r maximal so that (1 − y)r v = 0. The resulting vector is fixed by Y , whence (a). (b) By (a). (c) By inspection rad KY is an ideal, maximal because its codimension in KY is 1. Being the kernel of the trivial representation, which is the only irreducible one by (b), it is the unique maximal left ideal; and similarly for right ideals. On each factor of a composition series for the left regular representation of KY on itself Y acts trivially by (b), hence rad KY acts as 0, so that rad KY is nilpotent. (d) By (b).  Lemma 81. For x ∈ X − {1}, write wxw = f (x)h(x)wg(x) with f (x), g(x) ∈ X and h(x) ∈ H. (a) f and g are permutations of X − {1}.  (b) (Xw)2 = Xw2 + x∈X−{1} h(x)Xwg(x). Proof. (a) If f (x) = 1, we get the contradiction xw ∈ B, while if f (x ) = f (x), we see that w−1 x−1 x w ∈ B, so that x = x. Similarly for g.  (b) (Xw)2 = Xw2 + x∈X−{1} Xf (x)h(x)wg(x). Here f (x) gets absorbed in X and h(x) normalizes X, whence (b).  Proof of Theorem 44. (a) By Lemma 80(b) the space of fixed points of U = X on V is nonzero. Since H normalizes U and is abelian, that space contains a nonzero vector v such that (a) and (b) in the definition of highest weight vector hold. Let Xwv = v1 . If v1 = 0 then (c) holds with μ = 0. If not, we replace v Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

13. REPRESENTATIONS CONTINUED

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by v1 . Then (a) and (b) of the definition still hold with wλ in place of λ, and by  Lemma 81(b) so does (c) with μ = λ(h(x)). Now to prove that KGv = KY v it is enough, because of the decomposition G = Y B ∪ wB, to prove that wv ∈ KY v. By the two parts of Lemma 81 we may write Xwv = μv, after some simplification, in the form  λ(wh−1 w)y(x)v = μv (∗) wv + x∈X−{1}

with y(x) = wxw−1 ∈ Y , whence our assertion. (b) Let V  = Kv and V  = rad KY · v. It follows from Lemma 80(c) that the / V  , fixed sum V = V  + V  is direct. Now assume there exists some v1 ∈ V , v1 ∈  by X. We may assume that v1 ∈ V and also that v1 is an eigenvector for H since H is abelian. We have Xwv1 = wY v1 = 0, since Y (1 − y) = 0 for any y ∈ Y . Thus v1 is a highest weight vector. By (a), V = KY v1 ⊆ KY V  = V  , a contradiction, whence (b). (c) By (b) an irreducible KG-module determines its highest weight (λ, μ) uniquely. Conversely, assume that V1 and V2 are irreducible KG-modules with highest weight vectors v1 and v2 of the same weight (λ, μ). Set v = (v1 , v2 ) ∈ V1 ⊕V2 and / V , since otherwise we could write v2 = cv + v  then V = KGv = KY v. Now v2 ∈  with c ∈ K and v ∈ rad KY · v and then projecting on V1 and V2 get that c = 0 and c = 1, a contradiction. Thus we may complete the proof as in the proof of Theorem 39(e).  Theorem 45. Let (λ, μ) be the highest weight of an irreducible KG-module V . (a) If λ = 1, then μ = 0. If λ = 1, then μ = 0 or −1. (b) Every weight as in (a) can be realized. Thus the number of possibilities is |H| + 1.  Proof. (1) Proof of (b). In KG let Hλ = h∈H λ(h−1 )h, then u = XHλ wX and v = XHλ . As in the proof of Theorem 44(a), a simple calculation yields  Xw(u + cv) = λ(h(x))u + cXwHλ X. x∈X−{1}

Here H acts, from the left, according to the characters wλ, λ, wλ on the respective terms. Thus if wλ taking c = 0. If wλ = λ,  = λ, we may realize the weight (λ, 0) by we take c = − λ(h(x)) instead. Finally if λ = 1, then λ(h(x)) = −1 and we get μ = −1 by taking c = 0. To achieve (λ, μ) in an irreducible module we simply take KG(u + cv) modulo a maximal submodule. (2) If dim V = 1, then V is trivial and (λ, μ) = (1, 0). Since X and Y are p-groups, they act trivially by Lemma 80(b), whence (2). (3) If dim V = 1, then λ determines μ. Write V = V  ⊕ V  as in the proof of Theorem 44(b). We have V  = 0. Since Y fixes V  it fixes some line in it, uniquely determined in V , by Theorem 44(b) with Y in place of X. Since Y clearly fixes wv, we conclude that wv ∈ V  . Projecting (∗) of the proof of Theorem 44(a) onto V  , we get  (∗∗) λ(wh−1 w) = μ whence (3). (4) Proof of (a). Combine (1), (2) and (3). Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms



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 Corollaries to Theorems 44 and 45. (a) If λ = 1, then x∈X−{1} λ(h(x)) = 0. The number of solutions n(h) of h(x) = h with h given is, modulo p, independent of h, in particular for each h is at least 1 (cf. Lemma 64, Step (1)). (b) The irreducible representation of weight (λ, μ) can be realized in the left ideal generated by XHλ wX + cXHλ with c = 1 for the trivial representation (1, 0), c = 0 otherwise. (c) If L ⊂ K is a splitting field for H, it is one for G. (d) dim V = |X| if (λ, μ) = (1, −1), dim V < |X| if not. (e) The number of p-regular conjugacy classes of G is |H| + 1.  Proof. (a) If λ = 1, then μ = 0 by Theorem 45(a), λ(h(x)) = 0  so that by (∗∗) above applied with λ replaced by wλ. Then n(h)λ(h) = 0 for every λ = 1. By the orthogonality relations for the characters on H (which are valid since p  |h|), we conclude that n(h), as an element of K, is independent of h. If  n(h) were 0 for some h, we would get |H| = n(h) = 0 (mod p), a contradiction. (b) Let V be an irreducible module whose dual V ∗ has the highest weight (λ, μ). We consider, as in Theorem 40, the isomorphism ϕ of V ∗ into the induced representation space of functions a : G → K such that a(yb) = a(y)λ∗ (b) for all y ∈ G, b ∈ B defined by (ϕf )(x) = f (xv + ) with v + a highest weight vector for V . Using the decomposition V = Kwv + + rad KX · wv + , we may define f + as in Lemma 73, prove that it is a highest weight vector, and that a+ = ϕf + is given on G by the equation of Lemma 74, with λ∗ in place  of λ. Converting functions to elements of KG in the usual way, a ∼ a(x)x, we see that a+ becomes the element of (b), whence (b). At the same time we see that V may be realized in the induced module Bλ∗ → G, as the unique irreducible submodule in case λ = 1, as one of two in case λ = 1. (c) By (b). (d) If μ = 0, then Xwv = 0, whence Y v = 0 and dim V < |X| by Theorem 44(a). Conversely if dim V < |X|, then the annhilator of v in KY contains Y by Lemma 80(d), so that μ = 0. (e) By a classical theorem of Brauer and Nesbitt [10] the number in question equals the number of irreducible KG-modules, hence equals |H| + 1 by Theorem 45(b).  Example. G = SL2 (q). Here |H| = q − 1, so that |H| + 1 = q. Remarks. (a) We see that the extra condition Xv = μv serves two purposes. First it distinguishes the smallest module ∼ (1, 0) from the largest (1, −1). Secondly, in the proof of the key relation KGv = KU − v it takes the place of the density argument (U − B dense in G) used in the infinite case. (b) The preceding development applies to a wide class of doubly transitive permutation groups (with B the stabilizer of a point, H of two points), since it depends only on the facts that H is abelian and has in B a normal complement U which is a p-Sylow subgroup of G. Now we consider groups of arbitrary rank. W (= Wσ ) will be given the structure of a reflection group as in Theorem 32 with Σ/R projected into Vσ and scaled down to a set of unit vectors, the corresponding root system. For each simple root a, we write Ya for X−a and choose wa in Xa , Ya  to represent wa in W . If w ∈ W Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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is arbitrary, we choose a minimal expression w = wa wb · · · as a product of simple reflections, and set w = wa w b · · · . Then w is independent of the minimal expression chosen. We postpone the proof of this fact, which could (and probably should) have been given much earlier, to the end of the section so as not to interrupt the present development. As a consequence we have: Lemma 82. If w = wa wb · · · is any minimal expression, then X w = Xw a · X b wb · · · . Proof. Since w = wa w b · · · this easily follows by induction on N (w) or by Appendix II.25.  We extend the earlier definition of highest weight vector by the new requirement: (c) X a wa v = μa v (μa ∈ K) for every simple root a. Theorem 46. Let G be a (perhaps twisted) finite Chevalley group (of arbitrary rank). (a) (b) (c) Same as (a) (b) (c) of Theorem 44. (d) Let Ha = H ∩ Xa , Ya . If (λ, μa ) is the highest weight of some irreducible module then μa = 0 if λ|Ha = 1 and μa = 0 or −1 if λ|Ha = 1. (e) Every weight as in (a) can be realized on some irreducible KG-module. Proof. We shall prove this theorem in several steps. (a1) There exists in V a nonzero eigenvector v for B. This is proved as in Theorem 44(a). (a2) If v is as in (a1), then so is v1 = X a w a v (a simple), unless it is 0. Proof of (a2). Let x be any element of U . Write x = xa xa with xa ∈ Xa and xa ∈ Xa , the subgroup of elements of U whose Xa components are 1. We recall that Xa and wa normalize Xa (see Appendix I.11). Thus xv1 = xa X a wa v = v1 , since U v = v. Since H normalizes Xa and is normalized by wa , we see that v1 is also an eigenvector for H.  (a3) Choose v as in (a1), then w ∈ W so that N (w) is maximal subject to v1 = X w wv = 0. Then v1 is a highest weight vector. Proof of (a3). By Lemma 82 and (a2), v1 is an eigenvector for B. Let a be any simple root. If w−1 a > 0, then N (wa w) = N (w) + 1 by Appendix II.19, so that X wa w wa w = X a wa X w w by Lemma 82, and X a wa v1 = 0 by the choice of w. If w−1 a < 0, then we may choose a minimal expression for w = wa wb · · · starting with wa . Then X a w a v1 = (X a wa )2 X b wb · · · v = μX a w a X b w b · · · v = μv1 .

(by Lemma 82) (with μ ∈ K by Lemma 81(b))

By (a1) and (a3) we have the first statement in (a). (a4) KGv = KU − v. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms



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 Proof of (a4). We have G = w0 G ⊆ w wU − B by Theorem 4. Thus it is − enough to show that each w fixes KU w, and for this we may assume w = wa with a simple. Assume y ∈ U − . Write y = ya ya as above, but using negative roots instead. Then ya and w a normalize Ya , so that wa yv = w a ya ya v ∈ U − w a ya v ⊆ KU − v by  Theorem 44(a) applied to Xa , Ya . (b1) If V = V  + V  with V  = Kv and V  = rad KU − v as before, then V  is fixed by every X a wa .  Proof of (b1). Write y = ya ya as before. Then X a w a yv = x∈Xa y(x)xwa v   with y(x) ∈ U − . Thus X a w a (y − 1)v = x∈Xa (y(x) − 1)xwa v ∈ V  . (b2) Proof of (b). If this is false, there exists v1 ∈ V  , v1 = 0, v1 fixed by X. As usual we may choose v1 as an eigenvector for H, and then by (b1) and (a3) also an eigenvector for each X a w a . Then, as before, V = KGv1 = KU − v1 ⊆ V  , a contradiction.  (c) Same proof as for Theorem 44(c). (d) By Theorem 45(a) applied to Xa , Ya . (e) Let π be the set of simple roots a such that μa = 0 and Wπ the corresponding subgroup of W . The  reader should have no trouble proving that the left ideal of KG generated by w∈Wπ w0 U Hλ wX w is an irreducible KG-module whose highest weight is (λ, μa ). Corollary. (a) A splitting field for H is also one for G. (b) Let V be irreducible, of highest weight (λ, μa ). Then dim V = |U | if λ = 1 and all μa = −1, dim V < |U | if not. (c) For each set π of simple roots, let Hπ be the group generated by all Ha (a ∈ π). Then the number of irreducible  KG-modules, or equivalently, of p-regular conjugacy classes of G is π |H/Hπ |. Proof. (a) Clear. (b) Write U w0 v = X w0 w0 v = X a wa X b wb · · · v, with w0 = wa wb · · · as in Lemma 82, and then proceed as in the proof of Corollary (d) to Theorems 44 and 45. (c) The given sum counts the number of possible weights (λ, μa ) according to  the set π of simple roots a such that μa = −1. Exercise. If G is universal, the above number is   (|Ha | + 1) = q(α), a simple

α simple

in the notation of Theorem 43. If in addition G is not twisted, then the number is q l . It remains to prove the following result used (inessentially) in the proof of Lemma 82. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

13. REPRESENTATIONS CONTINUED

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Lemma 83. (a) If w ∈ W , then any two minimal expressions for w as a product of simple reflections can be transformed into each other by the relations (∗)

wa wb wa · · · = wb wa wb · · ·

(n terms on each side, n = order of wa wb , with a and b distinct simple roots). (b) Assume that for each simple root a the corresponding element wa of G (any Chevalley group) is chosen to lie in Xa , X−a . Let w = wa wb · · · be a minimal expression for w ∈ W . Then w = wa w b · · · is independent of the minimal expression chosen. Proof. (a) This is a refinement of Appendix IV.38 since the relations wa2 = 1 are not required. It is an easy exercise to convert the proof of the latter result into a proof of the former, which we shall leave to the reader. (b) Because of (a) we only have to prove (b) when w has the form of the two sides of (∗). For this we can refer to the proof of Lemma 56 since the extra restrictions there, that G is untwisted and that wa = wa (1) for each a, are not essential for the proof.  Remark. It would be nice if someone could incorporate in the elementary development just given the tensor product theorem mentioned after Theorem 43 or at least a proof that every irreducible K-module for G can be extended to the including algebraic group, hence also to other finite Chevalley groups contained in the latter group.

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Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

https://doi.org/10.1090//ulect/066/15

CHAPTER 14

Representations Completed Now we turn to the complex representations of the finite groups just considered. Here the theory in is poor shape. Only GLn (see Green [30]) and a few groups of low rank have been worked out completely, and then only in terms of the characters. Here we shall consider a few general results which may lead to a general theory. Henceforth K will denote the complex field. Given a (1-dimensional) character λ on a subgroup B of a group G, realized on a space Vλ , we shall write VλG for the induced module for G. This may be defined by VλG = KG ⊗KB Vλ (this differs from our earlier version in that we have not switched to a space  of functions), and may be realized in KG in the left ideal generated by Bλ = b∈B λ(b−1 )b (and will be used in this form). Its dimension is |G/B|. Exercise. Check these assertions. Lemma 84. Let B, C be subgroups of a finite group G, let λ, μ be characters on B, C, and let VλG , VμG be the corresponding modules for G. (a) If x ∈ G, then Bλ × Cμ in KG is determined up to multiplication by a nonzero scalar by the (B, C) double coset to which x belongs. (b) HomG (VλG , VμG ) is isomorphic as a K-space to the one generated by all Bλ × C μ . (c) If B = C and λ = μ, then the isomorphism in (b) is one of algebras. (d) The dimension of HomG (VλG , VμG ) is the number of (B, C) double cosets D such that Bλ × Cμ = 0 for some, hence every x ∈ D, or, equivalently, such that the restrictions of λ and xμ to B ∩ xCx−1 are equal. Proof. (a) This is clear. (b) Assume T ∈ HomG (VλG , VμG ). Since Bλ generates VλG as a KG-module,  T Bλ determines T . Let T Bλ = x∈G/C cx × Cμ . Since bBλ = λ(b)Bλ , we get by  averaging over B that T Bλ = x∈B\G/C cx Bλ × Cμ . Thus T is realized on Bλ ,  hence on all of VλG , by right multiplication by |B|−1 cx Bλ × Cμ . Conversely, any such right multiplication yields a homomorphism, which proves (b). (c) By discussion in (b). (d) The first statement follows from (a) and (b). Let B1 = B ∩ xCx−1 and C1 = x−1 Bx∩C, and {yi } and {zj } systems of representatives for B1 \B and C/C1 .   Set Bλ = λ(yi−1 )yi and Cμ = μ(zj−1 )zj . Then Bλ × Cμ = Bλ B1λ B1,xμ xCμ with (xμ)(xcx−1 ) = μ(c) since xC1 x−1 = B1 . If λ = xμ on B1 , then B1λ B1,xμ = 0. If λ = xμ, this product is |B1 | B1λ , and then Bλ × Cμ = 0 since the elements  yi b1 × zj are all distinct. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 137

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Remarks. (a) This is a special case of a theorem of G. Mackey (see, e.g., Feit [25]). (b) The algebra of (c) is also called the commuting algebra since it consists of all endomorphisms of VλG that commute with the action of G.48 Theorem 47. Let G be a (perhaps twisted) finite Chevalley group. (a) If λ is a character on H extended to B in the usual way, then VλG is irreducible if and only if wλ = λ for every w ∈ W such that w = 1. (b) If λ, μ both satisfy the condition of (a), then VλG is isomorphic to VμG if and only if λ = wμ for some w ∈ W . Proof. (a) VλG is irreducible if and only if its commuting algebra is 1-dimensional (Schur’s Lemma), i.e., by (c) and (d) of Lemma 84, if and only if λ and wλ agree on B ∩ wBw−1 , hence on H, for exactly one w ∈ W , i.e., for only w = 1. (b) Since VλG and VμG are irreducible, they are isomorphic if and only if dim HomG (VλG , VμG ) = 1, which, as above, holds exactly when λ = wμ for some (hence exactly one) w ∈ W .  Exercise. (a) dim HomG (VλG , VμG ) = |Wλ | if λ = wμ for some w, and equals 0 otherwise. (b) In Theorem 47 the conclusion in (b) holds even if the condition in (a) doesn’t. Here Wλ is the stabilizer of λ in W . We see, in particular that if λ = 1 then the commuting algebra of VλG is |W |-dimensional. But more is true. Theorem 48. Let V be the KG-module induced by the trivial 1-dimensional KB-module. Then the commuting algebra EndG (V ) is isomorphic to the group algebra KW . Reformulations. (a) The multiplicities of the irreducible components of V are just the degrees of the irreducible KW -modules. (b) The subalgebra, call it A, of KG spanned by the double coset sums BwX w is isomorphic to KW . (c) The algebra of functions f : G → K biinvariant under B (f (bxb ) = f (x) for all b, b ∈ B) with convolution as multiplication is isomorphic to KW . Proof. The theorem is equivalent to (a) by Schur’s Lemma and to (b) by Lemma 84, while (b) and (c) are clearly equivalent. We shall give a proof of (b), due to J. Tits. w

will denote the average in KG of the elements of the double coset BwB. The elements w

form a basis of A and 1 is the unit element. If a is a simple root, ca will denote |Xa |−1 . (1) A is generated as an algebra by {w

a : a simple} subject to the relations 2

(α) w

a = ca 1 + (1 − ca )w

a for all a. (β) w

a w

b w

a · · · = w

b w

a w

b · · · , as in Lemma 83(a). 48 Such

endomorphisms are also called intertwining operators. ()

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Proof of (1). We observe that if each ca is replaced by 1 then these relations go over into a defining set for KW , by Appendix II.38. Since B ∪Bwa Xa is a group, we have (Bwa X a )2 = rB + sBwa X a with r, s ∈ K. Since Bwa X contains with each of its elements its inverse, we get r = |Bwa Xa |, and then from the total coefficient s = |Bwa Xa | − |B|. Thus (α) holds in A, and so does (β) by Lemma 25, Corollary, which also shows that the w

a generate A. Conversely, let A1 be the abstract associative algebra (with 1) generated by symbols w

a subject to (α) and (β). For each w ∈ W choose a minimal expression w = wa wb · · · and set

b · · · . By Lemma 83(a) and (β) this is independent of the expression w

= w

a w chosen. By (α) it follows that ® if w−1 a > 0, w ‘ a w, w

a w

= ‘

if w−1 a < 0. ca w a w + (1 − ca )w, Thus the w

form a basis for A1 , which, having the same dimension as A, is therefore isomorphic to it.  (2) There exists a positive number c = c(G) such that ca = cna with na a positive integer depending only on the type of G. The multiplication table of A in terms of the basis {w}

is given by polynomials in c depending only on the type. Proof of (2). Consider 2 A4 (q 2 ), for example. Here the two possibilities for |Xa | are q 2 and q 3 by Lemma 63(c). If we set c = q −1 , then the corresponding values of na are 2 and 3, which depend only on the type. For each of the other types the verification is similar. From the first statement of (2) and the equations of the proof of (1) the second statement follows.  (3) An associative algebra Ac with multiplication table given by the polynomials of (2) exists for every complex number c. In particular, Ac(G) = A and A1 = KW . Proof of (3). Since the type of the group G contains an infinite number of members, the multiplication table is associative for an infinite set of values of c, hence for all values.  (4) Ac is semisimple for c = c(G), for c = 1, and for all but a finite number of values of c. Proof of (4). Ac is for c = c(G) the commuting algebra of a KG-module and for c = 1 a group algebra KW , hence semisimple in both cases. The discriminant of Ac is a polynomial in c, nonzero at c = 1 since then Ac is semisimple, hence nonzero for all but a finite number of values of c.  (5) Completion of the proof. Since A is semisimple and K is an algebraically closed field, A is a direct sum of complete matrix algebras, of certain degrees over K (see, e.g., Jacobson [34] or Feit [25]), and similarly for KW . We have to show that the degrees are the same in the two cases. If A is any finite-dimensional associative algebra which is separable, i.e., which is semisimple when the base field is extended to its algebraic closure, we define the numerical invariants of A to be the degrees of the resulting matrix algebras. The proof of Theorem 48 will be completed by the following lemma. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Lemma 85. Let R be an integral domain, F its field of quotients, and f a homomorphism of R onto a field K. Let A be a finite-dimensional associative algebra over R, and AF and AK the resulting algebras over F and K. If AF and AK are separable, then they have the same numerical invariants. In fact, from (3) and the lemma with R = K[c], A = Ac , and first f : c → c(G) and then f : c → 1, it follows that A and KW have the same numerical invariants, hence that they are isomorphic.  Proof of the lemma. (a) Assume that B is a finite-dimensional semisimple associative algebra over an algebraically closed field L, that b1 , b2 , . . . , bn form a basis  for B/L, that x1 , x2 , . . . , xn are independent indeterminates over L, that b = polynomial of b acting from the left on xi bi , and that P (t) is the characteristic BL(x1 ,...,xn ) , written as P (t) = Pi (t)pi with the Pi distinct monic polynomials irreducible over L(x1 , . . . , xn ). Then: (a1) The pi are numerical invariants of B. (a2) pi = degt Pi for each i. (a3) If P (t) = Qj (t)qj is any factorization over L(x1 , . . . , xn ) such that qj = degt Qj for each j, then it agrees with the one above so that the qj are the numerical invariants of B. Proof of (a). For (a1) and (a2) we may assume that B is the complete matrix  xij Eij in terms of the matrix units Eij . If X = [xij ], algebra End Lp and that b = then P (t) = det(tI − X)p , so that we have to show that det(tI − X) is irreducible over L(xij ). This is so since specialization to the set of companion matrices ⎤ ⎡ 1 ⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ .. ⎥ ⎢ . ⎥ ⎢ ⎣ 1 ⎦ xp1

xp2

···

xpp

yields the general equation of degree p. In (a3) if some Qj were reducible or equal to some Qk with k = j, then any irreducible factor Pi of Qj would violate (a2).  (b) Let R∗ be the integral closure of R in F (consisting of all elements satisfying monic polynomial equations over R), and x1 , . . . , xn indeterminates over F . Then R∗ [x1 , . . . , xn ] is the integral closure of R[x1 , . . . , xn ] in F (x1 , . . . , xn ). Proof of (b). See, e.g., Bourbaki, Commutatitive Algebra [8, Chapter V, Prop. 13].  (c) If R∗ is as in (b), then any homomorphism of R into K can be extended to one of R∗ into K. Proof of (c). By Zorn’s lemma this can be reduced to the case R∗ = R[α]  where it is almost immediate since K is algebraically closed. (d) Completion of proof. Let {ai } be a basis for A/R, hence also for AF /F , and {xi } independent indeterminates over F and also over K. The given homomorphism f : R → K defines a homomorphism f : A → AK . By (c) it extends to a homomorphism of R∗ into K and then naturally to one of R∗ [x1 , . . . , xn ] into Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

14. REPRESENTATIONS COMPLETED

141

 K[x1 , . . . , xn ]. If a = xi ai and P (t) = Pi (t)pi is its characteristic polynomial, factored over F (x1 , . . . , xn ) as before, then the coefficients of each Pi are integral polynomials in its roots, hence integral over the coefficients of P , hence integral  over R[x1 , x2 , . . . , xn ], hence belong to R∗ [x1 , . . . , xn ] by (b). Thus if f (a) = xi f (ai ), then its characteristic polynomial has a corresponding factor ization Pf (t) = Pif (t)pi over K(x1 , . . . , xn ). By (a1) the pi are the numerical invariants of AF , and by (a2) they satisfy pi = degt Pi = degt Pif , so that by (a3) with B = AK they are also the numerical invariants of AK , which proves the lemma.  Exercise. If λ is a character on H extended to B in the usual way, then EndG (VλG ) is isomorphic to KWλ . (Observe that this result includes both Theorem 47 and Theorem 48.) Remark. Although A is isomorphic to KW there does not seem to be any natural isomorphism and no one has succeeded in decomposing the module V of Theorem 48 into its irreducible components, except for some groups of low rank. We may obtain some partial results, in terms of characters, by inducing from the parabolic subgroups and using the following simple facts. Lemma 86. Let π be a set of simple roots, Wπ and Gπ the corresponding subgroups of W and G (see Lemma 30), and VπW and VπG the corresponding trivial modules induced to W and G; and similarly for π  . (a) A system of representatives in W for the (Wπ , Wπ ) double cosets becomes in G a system of represenatives for (Gπ , Gπ ) double cosets. (b) dim HomG (VπG , VπG ) = dim HomW (VπW , VπW ). Proof. (a) Exercise. (b) By Lemma 84(d) and (a).



of VπG and similarly for W . If Corollary 1. Let χG π denote the character  W W {nπ } is a set ofintegers such that χ = nπ χπ is an irreducible character of W , then χG = nπ χG π is, up to sign, one of G. Proof. Let (χ, ψ)G denote the average of χψ over G. We have G G G (χG π , χπ ) = dim HomG (Vπ , Vπ ),

and similarly for W . Since χW is irreducible, (χW , χW )W = 1, it follows that (χG , χG )G = 1, so that ±χG is irreducible, by the orthogonality relations for finite group characters.  Remarks. (a) In a beautiful paper in Berliner Sitzungsberichte, 1900, Frobenius [26] has constructed a complete set of irreducible characters for the symmetric group Sn , i.e., the Weyl group of type An−1 , as a set of integral combinations of the characters χW π . Using his method and the preceding corollary one can decompose the character of V in Theorem 48 in case G is of type An−1 (see Steinberg [57]). (b) The situation of (a) does not hold in general. Consider, for example, the group W of type B2 , i.e., the dihedral group of order 8. It has five irreducible modules (of dimensions 1,1,1,1,2), while there are only four χW π ’s to work with. (c) A result of a general nature is as follows. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Corollary 2. If the notation is as above and (−1)π is as in Lemma 66(d),  G then χ = (−1)π χG π is an irreducible character of G and its degree is |U |.  Proof. Consider χW = (−1)π χW π . By (8) in the proof of Theorem 26 on page 83, extended to twisted groups (check this using the hints given in the proof Hence ±χG is also one by of Lemma 66), χW = det, an irreducible  character. G π Corollary 1 above. We have χ (1) = (−1) |G/Gπ |. If G is untwisted and the base field has then by Theorem 4 applied to G and to Gπ this can be  q elements, π continued (−1) W (q)/Wπ (q) = q N = |U |, as in (4) of the proof of Theorem 26. If G is twisted, the proof is similar.  We continue with some remarks on the algebra A of Theorem 48(b). Lemma 87. The homomorphisms of A onto K are given by: f (w

a ) = 1 or −ca for each simple root a, subject to the condition that f (w

a ) is constant on each W -orbit. Proof. For a and b simple, let n(a, b) denote the order of wa wb in W . We claim that (∗) a and b belong to the same W -orbit if and only if there exists a sequence of simple roots a = a0 , a1 , . . . , ar = b such that n(ai , ai+1 ) is odd for every i. The equation (wai wai+1 )n = 1 with n odd can be rewritten to show that wai and wai+1 are conjugate, so that if the sequence exists then a and b are conjugate. If a and b are conjugate, then so are wa and wb , and this remains true when we project into the reflection group obtained by imposing on W the additional relations (wc wd )2 = 1 whenever n(c, d) is even. In this new group wa and wb must belong to the same component,49 so that the required sequence exists.

b ) whenever By (∗) the condition of the lemma holds exactly when f (w

a ) = f (w n(a, b) is odd, i.e., exactly when f preserves the relations (β) (of the proof of Theorem 48). Since f (w

a ) = 1 or −ca exactly when f preserves (α) (solve the quadratic), we have the lemma.  Remark. By finding the annihilator in A of the kernel of each of the homomorphisms of Lemma 87, we get a 1-dimensional ideal I in A. This corresponds to an irreducible submodule of multiplicity 1 in V , realized in the left ideal KGI of KG. By working out the corresponding idempotent, the degree of the submodule can be found.  Exercise. (a) If f (w

a ) = 1 for all a, show that I = K qw w

with qw = |Xw |, and the corresponding KG-module is the trivial one. (Hint: by writing W as a union

a − 1)(w

a + ca ) = 0, of right cosets relative to {1, wa } and writing (α) in the form (w show that I is as indicated.) 

and that in this case (b) If f (w

a ) = −ca for all a, show that I = K (det w)w, −1 2 = q the dimension is |U |. (Hint: if e is the given sum and c w w , show that e = me  with m = cw = |G/B| / |U |.) (c) For G = B2 (q) work out all (four) cases of Lemma 87, hence obtain the degrees of all (five) irreducible components of V . (d) Same for A2 and for G2 . 49 Looking at Dynkin diagram we see that the group W  is isomorphic to the group generated by reflections, so that the notion of an indecomposable component makes sense; moreover, wa = 1, wb = 1. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

14. REPRESENTATIONS COMPLETED

143

Remark. It can be shown that the module of (b) is isomorphic to the one with character χG as in Lemma 86, Corollary 2. If we reduce the element  (det w) |U/Xw | BwX w (which is |B| |U | e) of I mod p, we see from the proof of Theorem 46(e) that the given module reduces to the one of that theorem for which λ = 1 and every μa = −1. This latter module can itself be shown to be isomorphic to the one in Theorem 43 for which λ, α = q(α) − 1 for every α. From these facts and Weyl’s formula the character of the original module can be found up to sign and the result used to 2 prove that the number of p-elements (of order a power of p) of G is |U | . For the details see Steinberg [61]. Finally, we should mention that the algebra A admits an involution given by

a for all a (which in case ca → 1 and A → KW reduces to w

a → 1 − ca − w w → (det w)w). The preceding discussion points to the following Problem: Develop a representation theory for finite reflection groups and use it to decompose the module V (or the algebra A) of Theorem 48. It is natural that in studying the complex representations of G we have considered first those induced by characters on B since for representations of characteristic p this leads to a complete set. In characteristic 0, however, this is not the case, as even the simplest case G = SL2 shows. One must delve deeper. Therefore, we shall consider representations of G induced by (1-dimensional) characters λ on U . We can not expect such a representation ever to be irreducible since its degree |G/U | is too large (larger than |G|1/2 ), but what we shall show is that if λ is sufficiently general then at least it is multiplicity-free. In other words, EndG (VλG ) is abelian, hence a direct sum of fields. (If λ is not sufficiently general, we can expect the Weyl group to play a role, as in Theorem 48.) Before stating the theorem, we prove two lemmas. Lemma 88. Let k be a finite field and λ a nontrivial character from the additive group of k into K ∗ . Then every character can be written uniquely λc : t → λ(ct) for some c ∈ k. Proof. The map c → λc is a homomorphism of k into its dual, and its kernel is clearly 0.  Lemma 89. For w ∈ W the following conditions are equivalent. (a) If a and wa are positive roots and one of them is simple, then so is the other. (b) If a is simple and wa is positive, then wa is simple. (c) w = w0 wπ for some set π of simple roots, with w0 as usual and wπ is the corresponding object of Wπ . There are 2l possibilities for w. Proof. (c) ⇒ (a) Because wπ maps π onto −π and (∗) permutes the positive roots with support not in π (same proof as for Appendix I.11). (a) ⇒ (b) Obvious. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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(b) ⇒ (c) Let π be the set of simple roots kept positive, hence simple, by w. We claim: (∗∗) if α > 0 and supp a  π, then wa < 0. Write a = b + c with supp b ⊆ π, supp c ⊆ Π − π. Then wa = wb + wc. Here wc < 0 by the choice of π, and supp wc  wπ ⊇ supp wb. Thus wa < 0. If a is a simple root not in π then wwπ a < 0 by (∗) and (∗∗), while if a is in π this holds by the definition of π. Thus  wwπ = w0 , whence (c). Theorem 49. Let G be a finite, perhaps twisted, Chevalley group and λ : U → K ∗ a character such that λ|Xa = 1 if a is simple, λ|Xa = 1 if a is positive but not simple. Then VλG is multiplicity-free. In other words, EndG (VλG ) is abelian, or, equivalently, the subalgebra A of KG spanned by the elements Uλ hwUλ (h ∈ H, w ∈ W ) is abelian.  Here Uλ = u∈U λ(u−1 )u, and we assume that the w are chosen as in Lemma 83(b). Remarks. (a) If a is not simple, then usually Xa ⊆ DU , so that the assumption λ|Xa = 1 is superfluous, but this is not always the case, e.g., for B2 or F4 with |k| = 2 or for G2 with |k| = 3. In these latter groups, there are other possibilities, which because of their special nature will not be gone into here. (b) The proof to follow is suggested by that of Gelfand and Graev [28], who have given a proof for SLn and announced a general result for the untwisted groups. T. Yokonuma [67] has also given a proof for these latter groups, but his details are unnecessarily complicated. Proof of Theorem 49. The fact that A is abelian will follow from the existence of an (involutory) antiautomorphism f of G such that (a) f U = U . (b) λf = λ on U . (c) For each  double coset U nU such that Uλ nUλ = 0, we have fn = n (here n ∈ N = Hw). For since f extended to KG and then restricted to A is an antiautomorphism and at the same time the identity (by (a), (b), (c)) it is clear that A is abelian. The existence of f will be proved in several steps. (1) If Uλ nUλ = 0 and n ∈ Hw, then w = w0 wπ for some set π of simple roots. Proof of (1). By Lemma 89 we need only prove that if a is simple and wa positive then wa is simple. Writing the first Uλ above with the Xwa component on the right, and the second with the Xa component on the left, we get Xwa,λ nXa,λ n−1 = 0. Since λ is nontrivial on Xa it is also so on Xwa , whence wa is simple by the assumption on λ, which proves (1).  The condition in (1) essentially forces the correct definition of f . We set a∗ = −w0 a. If a is simple, so is a∗ . In order to simplify the discussion in one or two spots we assume henceforth that G (i.e., its root system) is indecomposable. If G is untwisted, we start with the graph automorphism corresponding to ∗ (see the corollary on page 91), compose it with the inversion x → x−1 , and finally with a diagonal automorphism so that the result f satisfies, not only f U = U but also λf = λ on U . This is possible because of Lemma 89 and the assumptions on λ in the theorem. If G is twisted, then we may omit the graph automorphism (because ∗ is then the identity), and use the explicit isomorphism Xa /DXa ∼ = k of (2) of Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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the proof of Theorem 36 in combination with Lemma 89 to achieve the second condition. We see that (2) f is an involutory antiautomorphism which satisfies the required conditions (a), (b). We must prove that it also satisfies (c). As consequences of the construction we have: (3) f h = w 0 hw−1 0 for every h ∈ H. (4) If a∗ = a, then f is the identity on Xa /DXa . (5) If a∗ = a, there exists a nontrivial element of Xa fixed by f . Proof of (5). For Xa of type A1 this follows from (4). For Xa of type 2 A2 we choose the element (t, u) of Lemma 63(c) with t = 2, u = 2 if p = 2, and t = 0, u = 1 if p = 2, since f (t, u) = (t, −ttθ u) (check this, referring to the construction of f ). For types 2 C2 and 2 G2 we may choose (0, 1) and (0, 1, 0) since f is the identity on {(0, u)} and {(0, u, 0)}.  (6) The elements w a ∈ G may be chosen so that: (6a) f wa = wa∗ for every simple root a. (6b) If a and b are simple and n ∈ N is such that nXa n−1 = Xb and λ(nxn−1 ) = λ(x) for all x ∈ Xa , then nwa n−1 = wb . Proof of (6). Under the action of f and the inner automorphisms in by elements of n as in (6b) the Xa (a simple) form orbits. From each orbit we select an element xa . If a∗ = a, we choose xa ∈ Xa as in (5), write it as (∗) xa = x1 wa x2 with x1 , x2 ∈ X−a , and choose wa accordingly. Since f is an antiautomorphism and fixes X−a , it also fixes wa by the uniqueness of the above form. If a∗ = a, we choose xa ∈ Xa , xa = 1, arbitrarily. We then use the equations f w a = w a∗ and in wa = w b of (6a) and (6b) to extend the definition of w to the orbit of a. We must show that this can be done consistently, that we always return to the same value. Let a1 , a2 , . . . , an be a sequence of simple roots such that a1 = an = a and for each j either aj+1 = a∗j or else there exists nj such that the assumptions in (6b) hold with aj , aj+1 , nj in place of a, b, n. Let g denote the product of the corresponding sequence of f ’s and inj ’s. We must show that g fixes wa . We have gXa = Xa , gX−a = X−a , and in fact g acts on Xa /DXa , identified with k, by multiplication by a scalar c as follows from the definition of f and the usual formulas for in . Since λg = λ by the corresponding condition on f and each in , it follows from Lemma 89 that c = 1, so that g is the identity on Xa /DXa . If DXa = 0, then g fixes the element xa above, hence also wa by (∗), whether g is an automorphism or an antiautomorphism. If DXa = 0, then G is twisted so that a∗ = a. If g is an automorphism, then by the proof of Theorem 36 from (5) on its restriction to Xa , X−a  is the identity so that it fixes w a , while if g is an antiautomorphism then by the same result its restriction coincides with that of f so that it fixes wa by the choice of w a .  Remark. If G is untwisted, the above proof is quite simple. We assume henceforth that the wa are as in (6). (7) If w = wa w b · · · as in Lemma 83(b) and w∗ = w0 w−1 w0−1 , then f w = w∗ . Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Proof of (7). Since wa wb · · · is minimal, · · · wb∗ wa∗ is also. (Check this.) Since f is an antiautomorphism it follows from (6a) that f w = · · · w b∗ wa∗ = · · · wb∗ wa∗ = w∗ .  (8) If w is as in (1) then f w = w. Proof of (8). w∗ = w in this case (see (7)).



(9) If n is as in (1) then f n = n. Proof of (9). By (1), n ∈ wH with w = w0 wπ . Assume a ∈ π. Then wa is simple and λ(nxn−1 ) = λ(x) for all x ∈ Xa , by the inequality in the proof of (1). Thus nwa n−1 = w wa by (6b), from which we get, on picking a minimal expression for wπ , that (∗) nwπ n−1 = wπ∗ . Since N (w) = N (w0 wπ ) = N (w0 ) − N (wπ ), it follows that if we put together minimal expressions for w and wπ we will get one for w0 . Thus w0 = w wπ by Lemma 83(b), and similarly w0 = wπ∗ w, so that (∗∗) w wπ w−1 = wπ∗ . If we now write n = wh, then h commutes with w π by (∗) and (∗∗). Hence fn = fh · fw = =

(since f is an antiautomorphism)

w0 hw−1 0 w w wπ hw−1 π

= wh = n.

(by (3) and (8)) (since w0 = w wπ ) (since h commutes with wπ ) 

Thus f satisfies condition (c) and the proof of Theorem 49 is complete.



Exercise. (a) Prove that if {w a } is as in (6) and w is as in (1), then Uλ wUλ = 0. (b) Deduce that if Hπ denotes the kernel of the set of simple roots π then the dimension of A,hence the number of irreducible components of VλG , in Theorem 49 is |Hπ |. Remark. The natural group for the preceding theorem seems to be the adjoint group extended by the diagonal automorphisms, a group of the same order as the universal group, but with something extra at the top instead of at the bottom. For this group, G , prove that the dimension above is just (|Ha | + 1) = q(α) in the notation of the exercise just before Lemma 83. Prove also that in this case VλG is independent of λ. Remark. The problem now is to decompose the algebra A of Theorem 49 into its simple (1-dimensional) components. If this were done, it would be a major step towards a representation theory for G. As far as we know this has been done only for the group A1 (see Gelfand and Graev [28, 29]). It would not, however, be the complete story. For not every irreducible G-module is contained in one induced by a character on U , i.e., by Frobenius reciprocity, contains a 1-dimensional U -module, Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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as the following, our final, example, due to M. Kneser, shows (although it is for some types of groups such as An ).50 As remarked earlier, reduction mod 3 yields an isomorphism of the subgroup W + of elements of determinant 1 of the Weyl group W of type E6 onto the group G = SO5 (3), the adjoint group of type B2 (3). If we reverse this isomorphism and extend the scalars we obtain a representation of G on a complex space V . The assertion is that U , i.e., a 3-Sylow subgroup of G, fixes no line of V . Consider the following diagram. α1

α5 α2

α4 α3 α6 α7

This is the Dynkin diagram of E6 with the lowest root α7 adjoined (α7 is the unique root in −D (see Appendix III.33), unique because all roots are conjugate in the present case. It is connected as shown because of symmetry and the fact that each proper subdiagram must represent a finite reflection group.) We choose as a basis for V the α’s with α3 omitted, a union of three bases of mutually orthogonal planes. w1 w2 acts as a rotation of 120◦ in the plane α1 , α2  and as the identity in the other two planes, and similarly for w4 w5 and w6 w7 . The group W + also contains an element permuting the three planes cyclically as shown, because of the conjugacy of simple systems and the uniqueness of lowest roots, and the four elements generate a 3-subgroup of W + . It is now a simple matter to prove that this subgroup fixes no line of V .

50 Srinivasan

[56] computes the characters of irreducible representations of the group Sp4 (q) for odd q. One of these representations (θ10 in the notation of [56]) does not contain trivial representation when restricted to the commutant DU (and even to the center C(U ) ⊂ DU ); this follows from the fact that the sum of values of the character on elements of C(U ) equals 0. Therefore, this representation does not contain 1-dimensional U -submodules. () Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

https://doi.org/10.1090//ulect/066/16

Appendix on Finite Reflection Groups The results (and some of the terminology in what follows) are motivated by the theory of semisimple Lie algebras, but no knowledge of the theory is assumed. The main results are starred. I. Preliminaries V will be a finite-dimensional real or rational Euclidean space. By a reflection (on V ) is meant a reflection in some hyperplane H. If α is a nonzero vector orthogonal to H, the reflection, denoted σα , is given by (1) σα ρ = ρ − 2(ρ, α)/(α, α) · α (ρ ∈ V ). We observe that σα is an automorphism of V , or order 2. A useful fact is: (2) If w is an automorphism of V , then wσα w−1 = σwα . To prove this, apply both sides to ρ ∈ V , then use (1) and the invariance of (·, ·) under w. Σ will denote a finite set of nonzero elements of V such that: (3) α ∈ Σ ⇒ −α ∈ Σ and kα ∈ / Σ if k = ±1. (4) α ∈ Σ ⇒ σα Σ = Σ. The elements of Σ will be called roots, and W will denote the group generated by all σα (α ∈ Σ). (5) Lemma. The restriction of W to Σ is faithful. For, each w ∈ W fixes pointwise the orthogonal complement of Σ. (6) Corollary. W is finite. (7) Examples. (a) If Σ is the root system of a semisimple Lie algebra over the complex field, then W is the corresponding Weyl group. (b) If W is any finite group generated by reflections, e.g., the group of symmetries of a regular solid, Σ may be taken as the set of unit normals to the hyperplanes in which reflections of W take place. (8) Definitions. A subset of roots is called a positive system if it consists of the roots which are positive relative to some ordering of V . (Recall that this involves the specification of a subset V + of V which is closed under addition and under multiplication by positive scalars and satisfies trichotomy.) A subset of roots, say Π, is a simple system if (a) Π is a linearly independent set, and (b) every root is a linear combination of the elements of Π in which the nonzero coefficients are either all positive or all negative. ∗ (9) Proposition. (a) Each simple system is contained in a unique positive system. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms 149

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(b) Each positive system contains a unique simple system. Hence simple systems exist. If Π is a simple system, then clearly the “all positive” roots in (8b) form the unique positive system containing Π, whence (a). Now let P be any positive system. Let Π be a subset of P which generates P under positive linear combinations and is minimal relative to this property. Then (∗)

α, β ∈ Π, α = β ⇒ (α, β) ≤ 0.

Assume  not, so that σα β = β − cα with c > 0. Assume σα β ∈ P so that σα β = cγ γ (γ ∈ Π, cγ ≥ 0). If cβ < 1, the last equation, written suitably, expresses β as a positive combination of the other elements of Π, a contradiction to the minimality of Π, while if cβ ≥ 1, it expresses 0 as a positive combination of elements of Π, hence of P , equally a contradiction. Similarly, −σα β ∈ P leads contradiction, whence (∗). Now a linear relation on Π may be written  to a aα α = bβ β with the two sums over disjoint parts of Π and aβ , bβ ≥ 0. Writing this as ρ = σ and using (∗), we get (ρ, ρ) = (ρ, σ) ≤ 0, whence ρ = 0 and then every aα = 0 because the α’s are all positive. Similarly every bβ = 0. Thus Π is independent, is a simple system. From the definition of a simple system any simple system contained in P consists of those elements of P which are not positive combinations of others, hence is uniquely determined by P . (10) Lemma. Let Π be a simple system and P the positive system containing Π. (a) α, β ∈ Π, α = β ⇒ (α, β) ≤ 0. (b) ρ ∈ P ⇒ there exists α ∈ Π so that (ρ, α) > 0.  cα α (α ∈ Π, cα ≥ 0) as in (8b), and then use 0 < (ρ, ρ) =  For (b) write ρ = cα (ρ, α). ∗ (11) Main Lemma. Let Π, P be as in (10) and α ∈ Π. Then σα α = −α and σα (P − {α}) = P − {α}.  Pick ρ ∈ P − {α}, ρ = cβ β (cβ ≥ 0, β ∈ Π). By (3) some cβ > 0 (β = α). Application of σα does not change this cβ . Hence σα ρ ∈ P − {α}. (12) Theorem. Any two simple (or positive) systems are conjugate under W . By (9) we need only consider two positive systems, say P and P  . We use induction on n = |P ∩ (−P  )|. If n = 0, P = P  . Assume n > 0. Then there is a root α simple relative to P such that α ∈ P ∩(−P  ). By (11), |σα P ∩ (−P  )| = n−1, whence |P ∩ −σα P  | = n − 1. By the inductive assumption σα P  is conjugate to P ; hence so is P  . Henceforth Π, P will be as in (10) and fixed.   (13) Definition. If ρ ∈ Σ, ρ = α∈Π cα α as in (8b), then cα is called the height of ρ and written ht ρ. e.g., α ∈ Π ⇒ ht α = 1. (14) Lemma. Let W0 be the group generated by {σα : α ∈ Π}. If ρ ∈ P , the minimum value of ht on the set W0 ρ ∩ P is 1 and is taken on only W0 ρ ∩ Π. Let ρ be a minimum point and assume, if possible, that ρ ∈ / Π. By (10b) there is α ∈ Π so that (ρ , α) > 0, whence by (1) ht σα ρ < ht ρ and by (11) σα ρ > 0, a contradiction to the choice of ρ . (15) Corollary. (a) If ρ ∈ P , ρ ∈ / Π, then ht ρ > 1. (b) W0 Π = Σ, i.e., Every root ρ is conjugate under W0 to a simple root. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

II. THE FUNCTION N

151

By (14), (a) is clear and so is (b) if ρ > 0. If ρ < 0, then −ρ > 0, whence −ρ = wα (w ∈ W0 , α ∈ Π), so that ρ = (wσα )α. (16) Theorem. W is generated by {σα : α ∈ Π}. i.e., W = W0 in (14). If ρ is a root we have ρ = wα (w ∈ W0 , α ∈ Π), by (15b), whence σρ = wσα w−1 (see (2)), an element of W0 . Hence W ⊆ W0 and W = W0 . II. The Function N (17) Definition. For w ∈ W , N (w) will denote& the number of& roots ρ such that ρ > 0 and wρ < 0. In other words, N (w) = &P ∩ w−1 (−P )&. e.g., N (1) = 0, N (σα ) = 1 if α ∈ Π, by (11). (18) Lemma. w ∈ W ⇒ N (w−1 ) = N (w). Prove this. (19) Lemma. Assume w ∈ W , α ∈ Π. (a) If w−1 α > 0, then N (σα w) = N (w) + 1. (a ) If w−1 α < 0, then N (σα w) = N (w) − 1. (b) If wα > 0, then N (wσα ) = N (w) + 1. (b ) If wα < 0, then N (wσα ) = N (w) − 1. Let S(w) = P ∩ w−1 (−P ). Then S(σα w) = w−1 α ∪ S(w), whence (a). To get (a ) replace w by σα w in (a), and to get (b) and (b ) replace w by w−1 and use (18). (20) Problem. (a) N (ww ) ≤ N (w) + N (w ) and N (ww ) = N (w) + N (w ) (mod 2). (b) det w = (−1)N (w) . (w, w ∈ W ) 

(21) Lemma. Assume w = w1 w2 · · · wn (wi = σαi , αi ∈ Π). If N (w) < n, then for some i, j (1 ≤ i ≤ j ≤ n − 1), we have (a) αi = wi+1 wi+2 · · · wj αj+1 . (b) wi+1 wi+2 · · · wj+1 = wi wi+1 · · · wj .

i · · · w

j+1 · · · wn , where w

k denotes that wk has been omitted. (c) w = w1 w2 · · · w By (19b) and N (w) < n, w1 w2 · · · wj αj+1 < 0 for some j ≤ n − 1. Since αj+1 > 0, we have wi (wi+1 · · · wj αj+1 ) < 0 and wi+1 · · · wj αj+1 > 0 for some i ≤ j, whence wi+1 · · · wj αj+1 = αi by (11), which is (a). Using (2) with w = wi+1 · · · wj and α = αj+1 , we get (b), and then replacing the left side of (b) by the right side in the product for w and using wi2 = 1, we get (c). Problem. Prove, conversely, that (a), (b) or (c) implies N (w) < n. (22) Corollary. If w ∈ W , then N (w) is the number of terms in a minimal expression of w as a product of reflections corresponding to simple roots. Let w = w1 w2 · · · wn be a minimal expression. By (19) ((a) or (b)), N (w) ≤ n, and by (21c), N (w) ≥ n. ∗ (23) Theorem. For w ∈ W , if wP = P or wΠ = Π or N (w) = 0, then w = 1. The three assumptions are clearly equivalent. Now N (w) = 0 implies that the minimal expression of w in (22) is empty, whence w = 1. ∗ (24) Theorem. W is simply transitive on the positive systems, and also on the simple systems. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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By (12) and (23). (25) Problem. (a) For w ∈ W , choose a minimal expression as in (22), w = w1 w2 · · · wn (wi = σαi , αi ∈ Π) (so that n = N (w)), and set ρi = w1 w2 · · · wi−1 αi . Prove that ρi (1 ≤ i ≤ n) is a complete list of all roots ρ such that ρ > 0 and w−1 ρ < 0. (b) Since −P is a positive system, there exists by (24) a unique w0 ∈ W such that w0 P = −P . Write w0 = w1 w2 · · · wn (n = N (w0 ) = |P |) as above. Prove that ρi (1 ≤ i ≤ n) is a complete list of all positive roots. (Hint: (19), (21).)

III. A Fundamental Domain for W (26) Definition. D will denote the region {ρ ∈ V : (ρ, α) ≥ 0, α ∈ Π}. Thus D is a closed convex cone, and if Π spans V it is even a simplicial cone with vertex at 0. (27) Lemma. Every ρ ∈ V is conjugate to some ρ ∈ D. In fact to some ρ in D such that ρ − ρ is a nonnegative combination of the elements of Π. Let S be this set of nonnegative combinations (in other words, the dual cone of D). We introduce a partial order in V by the definition γ ≥ γ  if and only if γ − γ  ∈ S. Among the conjugates ρ of ρ under W such that ρ ≥ ρ, we pick one which is maximal relative to this partial order. Then α ∈ Π ⇒ σα ρ  ρ ⇒ (ρ , α) ≥ 0 (by (1)), whence ρ ∈ D and (27) follows. (28) Theorem. Assume w ∈ W, ρ ∈ V , ρ not orthogonal to any root, and wρ = ρ. Then w = 1. (Restatement: w = 1, wρ = ρ ⇒ ρ is orthogonal to some root.) By (27) we may assume ρ ∈ D. For α ∈ P , (wα, ρ) = (α, w−1 ρ) = (α, ρ) > 0. Hence wα ∈ P , for all α ∈ P , so that wP = P . Then w = 1 by (23). (29) Corollary. If ρ ∈ V is not orthogonal to any root, its conjugates under W are all distinct. (And conversely, of course.) (30) Corollary. The only reflections in W are those in hyperplanes orthogonal to roots, i.e., those of the form σρ (ρ ∈ Σ). Let u be any reflection in a hyperplane H not orthogonal to any root. The roots being finite in number, there exists ρ ∈ H, ρ not orthogonal to any root. Then u = 1, uρ = ρ ⇒ u ∈ / W , by (28). (31) Problem. Let S be a set of roots such that {σα : α ∈ S} generates W . Prove that every root is conjugate, under W , to some α ∈ S, and every reflection in W to some σα (α ∈ S). (32) Lemma. Assume ρ, σ ∈ D, w ∈ W , wρ = σ. Then (a) w is a product of simple reflections (i.e., relative to simple roots) fixing ρ. (b) ρ = σ. For (a) we use induction on N (w). If N (w) = 0, then w = 1 by (23). Assume N (w) > 0. Pick α ∈ Π so that wα < 0. Then 0 ≥ (σ, wα) = (ρ, α) ≥ 0, whence (ρ, α) = 0 and σα ρ = ρ. Since (wσα )ρ = σ, and N (wσα ) = N (w) − 1 by (19b ), the inductive assumption applied to wσα yields (a). Clearly (a) implies (b). ∗ (33) Theorem. D is a fundamental domain for W on V . In other words, each element of V is conjugate to exactly one element of D. By (27) and (32b). (34) Problem. If ρ ∈ D and w ∈ W , show that ρ−wρ is a nonnegative combination of positive roots. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

IV. GENERATORS AND RELATIONS FOR W

153

(35) Restatement. The reflecting hyperplanes (those orthogonal to roots) partition V into closed chambers, each of which is a fundamental domain for W . For a given chamber, the roots normal to the walls and inwardly directed form a simple system, and each simple system is obtained in this way. Prove these assertions and also that the angle between two walls of a chamber is always a submultiple of π. ∗ (36) Theorem. If S is any subset of V , the subgroup of W which fixes S pointwise is a reflection group. In other words, every w ∈ W which fixes S pointwise is a product of reflections which also do. Remark. (36) is an extension of (28). Verify this. For the proof of (36) we may assume that S is independent, hence finite, and using induction, reduce to the case where S has a single element, say ρ, which may be taken in D by (27). Then (32a) with σ = ρ yields our result. (37) Problem. For each Π ⊆ Π let W (Π ) denote the group generated by {σα : α ∈ Π }. Prove that W (Π ∩ Π ) = W (Π ) ∩ W (Π ).

IV. Generators and relations for W ∗ (38) Theorem. For α, β ∈ Π, let n(α, β) denote the order of σα σβ in W . (So n(α, α) = 1, while n(α, β) > 1 if α = β.) Then  the group W is defined by  the generators {σα : α ∈ Π} subject to the relations (σα σβ )n(α,β) = 1 : α, β ∈ Π . In other words, the given elements generate W and the given relations imply all others in W . By (16) the given elements generate W . Suppose the relation (∗)

w1 w2 · · · wr = 1,

(wi = σαi , αi ∈ Π),

holds in W . We will show it is a formal consequence of the given relations, by induction on r. By (20b) or by (19) r is even, say r = 2s. If s = 0 there is nothing to show. Suppose s > 0. we start with the observation: (0) (∗) is equivalent to wi+1 wi+2 · · · wr w1 w2 · · · wi = 1 (1 ≤ i ≤ r). Case 1. Suppose α1 = w2 w3 · · · ws αs+1 . We have N (w1 w2 · · · ws+1 ) = N (w2s w2s−1 · · · ws+2 ) < s + 1 by (19a). Hence by (21) we have (21a) and (21b) for some i, j such that 1 ≤ i ≤ j ≤ s. Since i, j = 1, s is excluded in the present case, both sides of (21b) have length < s. By our inductive assumption we may replace the left side of (21b) by the right side in (∗). If wi2 is then replaced by 1, the inductive assumption can be applied to the resulting relation to complete the proof, in the present case. Case 2. Suppose α1 = α3 . (If s = 1, this case doesn’t occur.) By the first case we may assume α1 = w2 w3 · · · ws αs+1 and then by (0) also α2 = w3 w4 · · · ws+1 αs+2 , whence (∗∗) w2 w3 · · · ws+1 = w3 w4 · · · ws+2 by (2). If (∗∗) is substituted into (∗), we can shorten (∗), as above. Thus we are reduced to showing that (∗∗) is a consequence of the original relations in (38), i.e., that w3 w2 w3 · · · ws+1 ws+2 ws+1 · · · w4 = 1 is. Since α3 = α1 = w2 w3 · · · ws αs+2 , we are back in Case 1. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Because of (0) above the only case that remains is: Case 3. Suppose α1 = α3 = α5 = · · · and α2 = α4 = α6 = · · · . Then (∗) has the form (σα σβ )s = 1 with α = α1 , β = α2 . Here s must be a multiple of n(α, β), the order of σα σβ , so that (∗) is a consequence of the relation (σα σβ )n(α,β) = 1. (39) Examples. (a) W = Sn . The symmetric group of degree n acts on an n-dimensional space by permuting the coordinates relative to an orthonormal basis i (1 ≤ i ≤ n). The transposition (i, j) corresponds to the reflection in the hyperplane orthogonal to i − j . We may take Σ = {i − j : i = j}, and relative to a lexicographic ordering, Π = {i − i+1 : 1 ≤ i ≤ n − 1}. Thus Sn is generated by the transpositions wi = (i i + 1) (1 ≤ i ≤ n − 1) subject to the relations wi2 = 1, (wi wi+1 )3 = 1, and (wi wj )2 = 1 if |i − j| > 1. (b) W = Octn . The octahedral group includes sign changes as well as permutations of the coordinates, so has order 2n n!. Here we may take Σ = {±i ± j , ±i : i = j} and Π = {i − i+1 , n : 1 ≤ i ≤ n − 1}. So, comparing with (a), we have one more generator wn and n more relations wn2 = 1, (wn−1 wn )4 = 1, (wi wn )2 = 1 if i ≤ n − 2. We observe that Sn and Octn are the groups of symmetries of the regular simplex and the regular cube. ∗ (40) Problem. Prove that W is defined by the generators {σα : α ∈ Σ} subject to the relations   (A) σα2 = 1 : α ∈ Σ ,   (B) σα σβ σα−1 = σγ : α, β ∈ Σ, γ = σα β . (Hint. Using (15b) and (16) show that the group so defined is generated by {σα : α ∈ Σ} as a consequence of (B), and then using (21) show that any nontrivial relation (wi = σαi , αi ∈ Π) w1 w2 · · · wn = 1 which holds in W can be shortened as a consequence of (A) and (B).) V. Appendix We consider some refinements in our results which occur in the crystallographic case, when 2(α, β)/(β, β) is an integer for all α, β ∈ Σ, which we henceforth assume. (This case occurs when we have root systems of Lie algebras.) (41) Refinement of (8). In the present case, all coefficients in (8b) are integers. Prove this, by induction on ht ρ (ρ ∈ Σ) (see (13)). (42) Corollary. ht ρ is always an integer. (43) Problem. Under the assumption of (34), assume also that (2ρ, α)/(α, α) is an integer for every α ∈ Σ. Show that ρ − wρ is a nonnegative integral combination of the elements of Π. (44) Problem. If α and β are roots, α = −β and (α, β) < 0, prove that α + β is a root. (Hint: prove that α + β equals σα β or σβ α.)

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Index

algebra commuting, 138 algebraic subset, 37 automorphism Frobenius, 123

groups sporadic, 114 highest weight, 15 highest weight vector, 15 homomorphism, 11 of algebraic groups, 37

Borel subgroup, 38 Cartan decompositioin, 66 Cartan integers, 5 central extension, 47 universal, 47 Chevalley basis, 7 Chevalley group, 19 closed set of roots, 21 complete algebaic variety, 95 covering, 47, 54 universal, 54 degree, 13 diagonal automorphism, 92 field automorphism, 93 flag incident, 27 standard, 26 graph isomorphism, 92 group adjoint, 30 fundamental, 30 Janko, 114 Mathieu, 114 McLaughlin, 115 Ree, 105 semisimple, 38 simply connected, 54 Suzuki, 115 Suzuzki, 105 universal, 30

ideal, 21 index, 100 isomorphism, 37 lattice, 15 matrix algebraic group, 37 maximal torus, 38 monomial, 13 radical, 38 representation contragredient, 16 linear, 49 projective, 49 root, 5 global, 39 Schur multiplier, 49 simplex, 27 string of roots, 6 subgroup Borel, 38 monomial, 26 parabolic, 34 unipotent, 21 subgroups commensurable, 72 system positive, 149 simple, 149 theorem Birkhoff-Witt, 11 Lie–Kolchin, 95 universal enveloping algebra, 11

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160

INDEX

weight, 14 fundamental, 28 weight vector, 14 Weyl group, 5 Weyl’s formula, 126 Whitehead group, 56 Zarisky topology, 37

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Selected Published Titles in This Series 66 Robert Steinberg, Lectures on Chevalley Groups, 2016 65 Alexander M. Olevskii and Alexander Ulanovskii, Functions with Disconnected Spectrum, 2016 64 Larry Guth, Polynomial Methods in Combinatorics, 2016 63 Gon¸ calo Tabuada, Noncommutative Motives, 2015 62 H. Iwaniec, Lectures on the Riemann Zeta Function, 2014 61 Jacob P. Murre, Jan Nagel, and Chris A. M. Peters, Lectures on the Theory of Pure Motives, 2013 60 William H. Meeks III and Joaqu´ın P´ erez, A Survey on Classical Minimal Surface Theory, 2012 59 58 57 56

Sylvie Paycha, Regularised Integrals, Sums and Traces, 2012 Peter D. Lax and Lawrence Zalcman, Complex Proofs of Real Theorems, 2012 Frank Sottile, Real Solutions to Equations from Geometry, 2011 A. Ya. Helemskii, Quantum Functional Analysis, 2010

55 Oded Goldreich, A Primer on Pseudorandom Generators, 2010 54 John M. Mackay and Jeremy T. Tyson, Conformal Dimension, 2010 53 John W. Morgan and Frederick Tsz-Ho Fong, Ricci Flow and Geometrization of 3-Manifolds, 2010 52 Marian Aprodu and Jan Nagel, Koszul Cohomology and Algebraic Geometry, 2010 51 J. Ben Hough, Manjunath Krishnapur, Yuval Peres, and B´ alint Vir´ ag, Zeros of Gaussian Analytic Functions and Determinantal Point Processes, 2009 50 John T. Baldwin, Categoricity, 2009 49 J´ ozsef Beck, Inevitable Randomness in Discrete Mathematics, 2009 48 Achill Sch¨ urmann, Computational Geometry of Positive Definite Quadratic Forms, 2008 47 Ernst Kunz, David A. Cox, and Alicia Dickenstein, Residues and Duality for Projective Algebraic Varieties, 2008 46 Lorenzo Sadun, Topology of Tiling Spaces, 2008 45 Matthew Baker, Brian Conrad, Samit Dasgupta, Kiran S. Kedlaya, and Jeremy Teitelbaum, p-adic Geometry, 2008 44 Vladimir Kanovei, Borel Equivalence Relations, 2008 43 Giuseppe Zampieri, Complex Analysis and CR Geometry, 2008 42 Holger Brenner, J¨ urgen Herzog, and Orlando Villamayor, Three Lectures on Commutative Algebra, 2008 41 James Haglund, The q, t-Catalan Numbers and the Space of Diagonal Harmonics, 2008 40 39 38 37

Vladimir Pestov, Dynamics of Infinite-dimensional Groups, 2006 Oscar Zariski, The Moduli Problem for Plane Branches, 2006 Lars V. Ahlfors, Lectures on Quasiconformal Mappings, Second Edition, 2006 Alexander Polishchuk and Leonid Positselski, Quadratic Algebras, 2005

36 Matilde Marcolli, Arithmetic Noncommutative Geometry, 2005 35 Luca Capogna, Carlos E. Kenig, and Loredana Lanzani, Harmonic Measure, 2005 34 E. B. Dynkin, Superdiffusions and Positive Solutions of Nonlinear Partial Differential Equations, 2004 33 Kristian Seip, Interpolation and Sampling in Spaces of Analytic Functions, 2004 32 Paul B. Larson, The Stationary Tower, 2004 31 John Roe, Lectures on Coarse Geometry, 2003 30 Anatole Katok, Combinatorial Constructions in Ergodic Theory and Dynamics, 2003

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Robert Steinberg’s Lectures on Chevalley Groups were delivered and written during the author’s sabbatical visit to Yale University in the 1967–1968 academic year. The work presents the status of the theory of Chevalley groups as it was in the mid-1960s. Much of this material was instrumental in many areas of mathematics, in particular in the theory of algebraic groups and in the subsequent classification of finite groups. This posthumous edition incorporates additions and corrections prepared by the author during his retirement, including a new introductory chapter. A bibliography and editorial notes have also been added. This is a great unsurpassed introduction to the subject of Chevalley groups that influenced generations of mathematicians. I would recommend it to anybody whose interests include group theory. —Efim Zelmanov, University of California, San Diego Robert Steinberg’s lectures on Chevalley groups were given at Yale University in 1967. The notes for the lectures contain a wonderful exposition of the work of Chevalley, as well as important additions to that work due to Steinberg himself. The theory of Chevalley groups is of central importance not only for group theory, but also for number theory and theoretical physics, and is as relevant today as it was in 1967. The publication of these lecture notes in book form is a very welcome addition to the literature. —George Lusztig, Massachusetts Institute of Technology Robert Steinberg gave a course at Yale University in 1967 and the mimeographed notes of that course have been read by essentially anyone interested in Chevalley groups. In this course, Steinberg presents the basic constructions of the Chevalley groups over arbitrary fields. He also presents fundamental material about generators and relations for these groups and automorphism groups. Twisted variations on the Chevalley groups are also introduced. There are several chapters on the representation theory of the Chevalley groups (over an arbitrary field) and for many of the finite twisted groups. Even 50 years later, this book is still one of the best introductions to the theory of Chevalley groups and should be read by anyone interested in the field. —Robert Guralnick, University of Southern California A Russian translation of this lecture course by Robert Steinberg was published in Russia more than 40 years ago, but for some mysterious reason has never been published in the original language. This book is very dear to me. It is not only an important advance in the theory of algebraic groups, but it has also played a key role in more recent developments of the theory of Kac-Moody groups. The very different approaches, one by Tits and another by Peterson and myself, borrowed heavily from this remarkable book. —Victor Kac, Massachusetts Institute of Technology For additional information and updates on this book, visit www.ams.org/bookpages/ulect-66

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